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Full text of "Practical plane and solid geometry for advanced students, including graphic statics, adapted to the requirements of the South Kensington syllabus"

NE and SOLID 



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i ADVANCED STUDENTS 



SON and BAXANDALL 



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PRESENTED 



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PRACTICAL 
PLANE AND SOLID GEOMETRY 




PRACTICAL 

PLANE AND SOLID 

GEOMETRY 

FOR ADVANCED STUDENTS 

INCLUDING GRAPHIC STATICS 

ADAPTED TO THE REQUIREMENTS OF THE 

ADVANCED STAGE OF THE SOUTH 

KENSINGTON SYLLABUS 

BY 

JOSEPH HARRISON 

M.I.M.E. , ASSOC. M. INST. C.E., 

INSTRUCTOR IN .MECHANICS AND MATHEMATICS AT THE ROYAL 
COLLEGE OF SCIENCE, LONDON 

AND 

G. A. BAXANDALL 

ASSISTANT IN MECHANICS AND MATHEMATICS AT THE ROYAL 
COLLEGE OF SCIENCE, LONDON 




Eonbott 
MACMILLAN AND CO., Limited 

NEW YORK : THE MACMILLAN COMPANY 
l8 99 



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PREFACE 

The training in Practical Geometry suitable for the student 
of Art consists of working problems mainly in Plane 
Geometry and Perspective. The former because decora- 
tive designs are largely based on geometrical figures ; and 
the latter because an artist constantly deals with appearances. 

On the other hand, the student of Science finds in 
Practical Geometry a powerful engine of calculation, which 
may often with great advantage be employed in preference 
to analytical processes. The principal use to him of plane 
geometry is that it enables him to make graphical or 
semi-graphical computations. Descriptive geometry is of 
great importance since it is required in problems which 
involve three dimensions in space ; and also because it 
shows how to exhibit the actual forms and dimensions of 
solid objects. 

This book is written for Science students. The 
necessity of accurate draughtsmanship is insisted on 
throughout. We describe how the drawing instruments 
may be set and their efficiency maintained. And the 
numerical answers appended to many of the examples 
should help to prevent any relapse into slovenly and in- 
accurate work on the part of those students who are apt to 
become lax. 

In Euclid's system of pure geometry the mechanical 



VI PRACTICAL GEOMETRY 

appliances and geometrical constructions allowed are 
severely restricted, the former being limited to a straight- 
edge, pencil, compasses, and a single plane surface on which 
to draw. A length may be transferred or a circle drawn by 
using the compasses, and a straight line may be drawn 
through two points. A line may not, however, be drawn to 
touch two circles without first finding the points of con- 
tact, although this adds nothing to the practical accuracy of 
the result. 

These restrictions, which form an important part of 
Euclid's scheme of logic, may yet be quite unsuitable in 
actual drawing. And in fact, if strictly adhered to, they 
are found to hamper the student in his work, to be wasteful 
of time, to lead to inaccuracies which may nullify the result, 
and even to render problems impossible of solution, the 
difficulties of which may otherwise be successfully over- 
come. In illustration see Example 2, page 139, on the 
motion of a slide valve driven by a Gooch link. 

It is true that in examinations in Practical Geometry 
the use of the tee- and set-squares has caused Euclid's 
restrictions to be somewhat relaxed, in that parallel and 
perpendicular lines are usually permitted to be drawn 
without construction. But does this go far enough ? Is 
not our science still in leading-strings, tied to the systems 
of the pure geometrician, its humble function being to 
illustrate his principles ? 

No doubt a student already familiar with the proofs of 
Euclid may derive much benefit from illustrating the more 
important propositions by carefully drawing the figures to 
scale. For example he may learn how small need be the 
errors which are introduced into graphical work. At various 
parts of Section I. such examples have been inserted. 



PREFACE vii 



But in the authors' view a fuller conception of the scope 
and province of Practical Geometry is needed, one which, 
seeing in it a powerful instrument of mathematical investi- 
gation, recognises the limitations as to accuracy and the 
practical requirements of the draughtsman, and permits 
and teaches the use of any mechanical devices where these 
are found to be useful and beneficial as, for example, the 
employment of transparent templates, to be adjusted by 
trial by hand, such as are described in connection with 
some of the problems and examples of this book. 

Chapter VI., which deals with the use of squared paper, 
and the plotting of curves from co-ordinates, should be 
found useful in view of the increasing recognition of the 
importance of this branch of the subject. 

In Chapter VII., where three planes of reference are 
used, the treatment is somewhat new. But the method 
has been tried and interests students, and is introduced 
with confidence. It is intended that a student shall read 
this chapter without much assistance from his teacher. 
Thus at the beginning of Descriptive Geometry he gets 
some very clear notions of projection and of the geometry 
of space, and is also trained to improvise models where 
these are helpful. 

The chapter on metric projection goes rather beyond 
the requirements of the advanced stage. But to have 
omitted any part would have left the subject very 
incomplete. 

The space at our disposal for the section on Graphics 
was very limited, but it is hoped that this part has not 
been unduly condensed. 

It is no doubt convenient to have the figures and 
corresponding descriptions arranged so as to avoid the 



viii PRACTICAL GEOMETRY 



necessity of turning over a leaf when referring to a 
diagram, but occasionally this could only be done at the 
expense of crowding or condensation. 

Examples bearing on the problems will generally be 
found following the latter or in close proximity thereto. 
The miscellaneous examples which close the chapters are 
mainly derived from the past examination papers of the 
Department of Science and Art, advanced stage, and by 
working these the student will be able from time to time 
to test his proficiency. 

In order to ensure the attainment of the desirable degree 
of accuracy in graphical work generally, the figures should 
not be made too small. If in some cases it may appear that 
this condition has been ignored, the reason must be attri- 
buted to want of space. The student, with drawing-paper 
at his disposal, cannot urge this excuse. 

July 1899. 



CONTENTS 



SECTION I. PRACTICAL PLANE GEOMETRY 



CHAPTER I 

hAGE 

General Introduction . , i 



CHAPTER II 
Similar Rectilineal Figures Areas . , 26 

CHAPTER III 
Triangles Circles and Lines in Contact . 50 

CHAPTER IV 

Conic Sections .... .70 

CHAPTER V 
Special Curves .... .110 

CHAPTER VI 
Co-ordinates Plotting on Squared Paper . 148 



PRACTICAL GEOMETRY 



SECTION II. PRACTICAL SOLID GEOMETRY, 
OR DESCRIPTIVE GEOMETRY 



CHAPTER VII 

PAGE 

Position in Space defined and exhibited . . 176 



CHAPTER VIII 
Fundamental Rules ok Projection . . . 193 

CHAPTER IX 
The Straight Line and the Perpendicular Plane . 204 

CHAPTER X 
The Oblique Plank ..... 230 

CHAPTER XI 
Horizontal Projection, or Figured Plans . . 274 

CHAPTER XII 
Plane and Solid Figurks in given Positions . . 292 

CHAPTER XIII 
The Projection of Curves and Curved Surfaces . 308 

CHAPTER XIV 
Tangent Planes to Surfaces . . . 34 



CONTENTS xi 



CHAPTER XV 

PAGE 

Surfaces in Contact ..... 368 



CHAPTER XVI 

Intersections of Surfaces, or Interpenetkations of 

Solids ....... 384 



CHAPTER XVII 
Cast Shadows ...... 406 

CHAPTER XVIII 
Metric Projection . . . . 442 

CHAPTER XIX 
Miscellaneous Problems . 462 



SECTION III. GRAPHICS 

CHAPTER XX 
Graphic Arithmetic ..... 4S6 

CHAPTER XXI 

Graphic Statics . , . 502 



xil PRACTICAL GEOMETRY 



APPENDIX I 

PAGE 

Science and Art Department Examination, May 1899 538 



APPENDIX II 

Definitions and Theorems ok Pure Solid Geometry 542 

INDEX .... . 549 



SECTION I 
PRACTICAL PLANE GEOMETRY 

CHAPTER I 

GENERAL INTRODUCTION 

1. Scope of subject. Practical geometry is the science 
which is concerned with applications of the problems of 
pure geometry to cases in which the geometrical figures 
require to be drawn to scale, often with great accuracy. 

In pure geometry we have a set of propositions, 
arranged in logical sequence, and deduced by strict reason- 
ing based on a few fundamental axioms and certain mathe- 
matical abstractions or conceptions which are defined. 
Hand sketches serve the purpose of representing the ideal 
figures. 

In practical geometry the conditions are different. The 
principal requirement is the careful drawing, to scale, of 
geometrical figures by means of mathematical instruments. 
More freedom is allowed in the use of instruments. Thus, 
by means of the edge of the drawing-board, by tec-square 
and set-squares, by parallel rulers, and by the clinograph, we 
draw parallel, perpendicular, and symmetrically disposed 
lines without any construction, such as would be required 
by Euclid. 



2 PRACTICAL PLANE GEOMETRY chap. 

2. Limits of accuracy. In pure geometry a point is 
defined as having position, but not magnitude, and a line 
has no breadth. Now it is evident that, in order to be 
visible, a point must be of a definite size, and a line must 
have breadth. Moreover, it is impossible to actually draw 
a line which shall be perfectly straight, or a circle, curve, 
or, in fact, any figure with absolute accuracy. Consequently 
the results we obtain by graphical construction are subject 
to unavoidable errors, and in practical geometry these con- 
ditions are recognised at the outset. 

In striving to approximate as nearly as may be to the 
ideal, and to obtain the greatest degree of accuracy war- 
ranted by the conditions of the particular problem under 
consideration, we fix working limits to the errors within 
which our results should be confined. 

Thus a point, pricked in the paper with the point of a 
needle, need not be more than -j-j^- of an inch in diameter 
in order to be readily seen by a person with ordinary eye- 
sight. A line drawn with a lead pencil of suitable lead 
and properly sharpened need- not be more than tt^-q- of an 
inch wide ; and with a good " straight-edge," no part of a 
line, of say one foot in length, need be out of the true 
alignment by so much as the -g-^-g- of an inch. The lines 
on an engine-divided scale should be correct in position to 
within xoVo f an inch, and with care we should be able to 
measure and set off distances to within -g^ of an inch. 
Set-squares are easily adjusted, so that the errors in the 
angles are kept within one minute or -^ of a degree. 

An expert draughtsman would be able to reduce these 
errors, if the circumstances were such as to warrant the 
increased care and expenditure of time that would be neces- 
sary. The limits given above are ordinary practical work- 
ing limits, and should be expected to be attained by every 
student in his work. In order to maintain this standard of 
accuracy, the student should have access to the tools 
necessary to keep his instruments in good working order, 
such as the trying-plane, oil-stone, etc. 



GENERAL INTRODUCTION 



3. Instruments. A list of the principal drawing instru- 
ments required by the student is now given, with a brief 
description of some of them. 

The drawitig-board should be of "imperial" size, 30" x 
22", and the tee-square of corresponding length. A "half- 
imperial " board in addition is very convenient. 

The 6o set-square may be 9" or 10" long, and the 45 
set-square about 6" long. 

A clhiograph is almost indispensable ; see next article 
for illustration. 

The pencils should be of pure quality, and of hard lead, 
say HH, or HHH. 

The compasses, if the best, will have knee-joints and 
needle-points firmly secured, and the dividers will have a 
fine screw adjustment. The inking-pens need not have 
jointed nibs. 

A pricker is requisite, and may be made by breaking 
a length of about 1" from the sharp end of a stout needle, 
and inserting it in the wood of a penholder, leaving V or 
so of the point projecting. 

A 6" semicircular protractor is preferable to the common 
6" rectangular one, as an instrument for measuring and 
setting off angles with accuracy ; but the best form is the 
circular protractor, and the diameter of this may be about 
6". It is desirable that angles should be capable of being 
read or marked off to within -Jg- of a degree. 

The scale is a very important instrument, and a special 
description of it is given in Art. 5. 

French curves may be used as templates when lining in 
curves which have first been carefully drawn freehand. 

Good cartridge drawing-paper with a smooth surface 
answers every purpose. Dratving-pins if used should have 
thin heads, but the latter always impede the movement of 
the squares. To overcome this, a good plan is to secure 
the corners of the paper with sealing-wax. The student 
should avoid inserting pins in his boards anywhere but 
near the edges. 



PRACTICAL PLANE GEOMETRY chap. 



4. Setting of instruments. The lead pencil should be 
of hard quality, and finished to a fine chisel edge with glass 
paper as illustrated in the figure opposite. The leads of 
the compasses may be similarly sharpened. 

The edges of the drawing-board and of the tee-square and 
set-squares should be trued up, and the angles of the latter 
made accurate by using a trying-plane. 

The dinograph is useful for drawing series of lines 
which are respectively parallel, perpendicular, or symmetric- 
ally inclined. See Example 6. 

Examples. 1. Draw a fine line, and on it step off with the 
dividers ten divisions each \" long ; and from the last point 
step off one division in a direction at right angles to the others. 
Join the first and last points, and through the other points draw 
lines parallel to this line. The lines so drawn will be nearly 
T V' apart, and should be so fine as to be each distinct. 

2. Repeat example I, taking the divisions each T V' instead of A". 

The lines will now be only -j-JV' a P ai 't, and should still be dis- 
tinct from one another. 

3. By using the tee-square and set-square draw two lines cross- 

ing one another at right angles. Then starting from any 
point on one of the lines, and using the 45 set-square, draw 
four lines in succession across the angular spaces, meeting 
respectively on the lines first drawn. The last point should 
coincide with the first point, the four lines last drawn forming 
a square. If this figure be drawn big enough, it gives a very 
severe test of the accuracy of the 45 set-square. 

4. Draw a circle of about 4" radius, and through its centre draw 

the six possible lines with the tee- and 6o set-squares. Prick 
off the twelve points where these lines intersect the circle, and 
test whether the twelve chords joining these points are of 
equal lengths. Also test whether the directions of four of the 
chords agree with the 45 set-square, and whether the 6o and 
45 set-squares can be arranged with two of their edges in 
contact, and one resting on the tee-square, so that the upper 
edge of the outer square shall coincide in succession witii the 
eight remaining chords. 

5. Describe a circle of say ii/' radius ; draw the circumscribing 

hexagon by using the tee-square and 6o set-square. Prick 
off the corners of this hexagon, and test whether all the sides 
are of equal length. Also try whether the nine lines joining 
the alternate and opposite points agree in direction with one or 
other of the edges of the 6o set-square. 



GENERAL INTRODUCTION 





6. Let any line XY be drawn and a point P taken anywhere on 
the drawing-paper. It is required to draw through P, (a) 
a line parallel to XY; (b) a line perpendicular to XY; (c) a 
line symmetrically inclined with XY. 

(a) Place the edge AB of the clinograph on the edge of the tee- 
square ; move the tee-square along the edge of the board and 
the clinograph along 'the edge of the tee-square, setting the 
edge DE to coincide with XY. Again move tee-square (if 
necessary) and clinograph until DE passes through P, when 
the required line may be drawn. 

(b) After setting the edge DE to coincide with XY, turn the 
clinograph over through a right angle so that BC rests on the 
edge of the tee-square ; the required line through P is then 
readily drawn. 

(C) Proceed as in (a) ; then turn the clinograph on to its other 
face, still keeping AB on the edge of the tee-square ; a line 
may then be drawn through /'such that this line and A' Fare 
symmetrical with regard to horizontal and vertical lines. 



6 PRACTICAL PLANE GEOMETRY chap. 

5. The Scale. The most useful scales for our work are 
those which are decimally subdivided. The subdivisions 
may extend along the whole length, or be confined to end 
main divisions, being technically known as " close " and 
"open" divided scales respectively. One close divided 
and two open divided scales can be got on one edge, so 
that if both faces be used, four different scales of the former 
kind and eight of the latter can be set off on one piece of 
boxwood, by using all the available space on it. 

The figure shows portions at the ends of one face of a 
boxwood scale 1 2" long, with four open divided scales of 1", 
|", \", and \" respectively, decimally subdivided. On the 
other face it would be very convenient to have four scales 
forming the set f ", ", T V, s\", or the set ", \'\ ", T V", 
also decimally subdivided. The student should be pro- 
vided with one or other of the set of eight scales thus 
described. 

The divisions of the scale are marked o, 1, 2, 3, . . ., but 
they may be read as o, 10, 20, 30, . . ., or o, 100, 200, 
300, . . ., etc., or as o, .1, .2, .3, . . ., or o, .or, .02, 
.03, . . ., or o, .001, .002, .003, . . ., and so on. 

If the divisions are read, o, 1, 2, 3, . . ., as marked, 
then the subdivisions represent figures in the first decimal 
place, and a fraction of a subdivision could be estimated 
by a figure in the second decimal place. Thus the distance 
PQ on the Y scale would be read 3.24. 

If the figures marked are read as tens, i.e. 10, 20, 30, 
. . ., the subdivisions become units, and the fraction of a 
subdivision is in the first place of decimals. Thus the 
distance PQ would now be 32.4. 

If the readings of the open divisions are 100, 200, 300, 
. ., PQ reads 327. If 1000, 2000, 3000, . . ., PQ is 
3240, and so on. 

While if the numbers at the open divisions stand for 
,i, .2, .3, . . ., then PQ represents .324. If the numbers 
are taken to mean .01, .02, .03, . . ., PQ is read equal to 
.0324, and so on. 



GENERAL INTRODUCTION 



Q 



J- 
























y 





/ 


2 




4 




6 




8 




10 


'8 
















o \ 


T 





, 








/ 




<LM.li 


i 1 i 1 i 1 i 1 i 


u 



'ID 



yz 



o 







^ 



yl 



^ 



Examples. 1. Measure the distance AB : 

(a) On the scale of i" to I unit. Ans. 2.37. 

(b) On the scale of i" to ioo units. Ans. 237. 

(c) On the scale of \" to 10,000 units. Ans. 47500. 
(if) On the scale of |" to o. I unit. Ans. 0.95. 

(e) On the scale of ' to .001 unit. /4.r. 0.00475. 
(/) On the scale of ^" to 10 units. Ans. 71.3. 
(g) On the scale of |" to .01 unit. Ans. 0.0317. 

Note. The respective scales must be applied directly to the line, 
and the number of units read off, without the interposition of dividers. 



2. Mark off lengths of- 
(a) 

{<) 
(d) 

w 



3.78 units on the scale of 1" to I unit. 
697 units on the scale of ^" to 100 units. 
0.913 unit on the scale of i" to o. 1 unit. 
0.001427 unit on the scale of j" to .0001 
0.092 unit on the scale of " to o. 1 unit. 



unit. 



Note. The lengths must be marked off with a pencil or pricker 
direct from the scale, the latter being applied to the paper. Dividers 
should not be used. 



3. Set off lengths representing : 

(a) 2.37 feet, to the scale of 1 inch to the foot. 

(b) 64,500 lbs. to the scale of \ inch to 10,000 lbs. 



PRACTICAL PLANE GEOMETRY chap. 



6. Ratio and proportion. When we speak of the ratio 
of one magnitude to another, say of A to B, written A : B 
or A/B, we signify how many times the first contains the 
second. 

Thus comparing a guinea with a crown, their respective 
money-values are in the ratio of 21 to 5, and this ratio 2 y x 
equals 4.2. Thus a ratio may be expressed as a pure 
number, that is, one not associated with kind or quantity. 

Or again the numbers on an ordinary scale give the 
ratios of the several lengths to the unit length. 

Proportion implies the equality of two ratios. Let the 
ratio A : B between two magnitudes be equal to the ratio 
a : b between two other magnitudes, the four terms A, B, a, b, 
are said to form a proportion. This equality is expressed 
in symbols thus 

A : B : : a : b, 
or A: B = a: b, 

A a 
B=b> 

and by algebra it is readily seen that 

B h r> a 1 

= --. or B : A : : b : a ; 
A a 

A B 
also = , or A : a : : B : : 

a b 
and A x b = a x B. 

Similar polygons afford typical illustrations of ratios and 
proportion. The student is acquainted with the funda- 
mental property of such figures, viz. : 

Theorem. In two similar rectilinear figures the ratios of 
pairs of corresponding tines are all equal. 

Thus in the similar triangles ABC, Abe we have 

AB BC CA 



Ab " be ' cA ' 
or AB : BC: CA = Ab : be : cA. 



GENERAL INTRODUCTION 





And in the similar figures OABCD, Oabcd the ratios 
AB BC CD DA OA OB OC OD 
cd ' ~da J ~0a' ~0b' ~Oc' 



ab ' be ' 

are all equal. 



Od 



The following relations are important. The student should verify 
them by inserting simple numbers for the letters. 

M N 



If 



then 



rind 



in 



p-M+q-N M N 
= or , 
p * m + q n m 7i 

p-M+q-m p-N+q->r 



'Jlf+s'tn r'J\T+s'n 

where p, q, r, and s are any numbers, positive or negative, whole or 
fractional. Thus in the figure, 

Ab Ac AB-Ab AC -Ac bB 

or 



_ Ac 
A~B = AC' : 



AB 



AC 



AB' 



cC 

AC' 



Examples. 1. Measure the sides of the triangles ABC, Abe to 
three significant figures on any suitable scale, say by applying 
the j" decimal scale direct to the figure ; and, by numerical 
calculation, verify the equality of the three ratios stated on p. 8. 

2. Verify the equality of the given ratios for the quadrilaterals by 
measuring the lengths of all the lines and performing the 
divisions by arithmetic. Calculate to three significant figures ; 
find the average of the eight values, and taking this as the 
correct value of the ratio, estimate the percentage error of the 
one most out of truth. 



IO 



PRACTICAL PLANE GEOMETRY chap. 



7. Problem. To divide a given line AB into any 
number of equal parts, say seven. 

(a) By construction (no figure). Draw AC at any 
angle, say 45, to AB, and, selecting any unit, step it off 
with the dividers seven times along A C, from A to D ; 
or applying the scale to AC, mark off, by means of pencil 
or pricker, seven equal divisions. Join DB, and through 
the other points of division on AD draw lines parallel to 
DB ; these intersect AB at the required points. 

Note. In choosing the unit, care should be taken that AD is neither 

too short nor too long, or the construction becomes "ill conditioned," 
and the points on AB are not determined with sufficient precision. 

(/<) By trial with the dividers. This is the method of 
trial and error and should be encouraged. It is a training 
in the estimation of lengths, and with practice the required 
division is quickly and accurately effected. 

(c) By the use of ruled (or squared) tracing-paper or 
ruled celluloid. Place the ruled tracing-paper as shown 
over AB, with the line marked o passing through A. Put 
the point of a pricker through the tracing-paper at A and 
turn the paper round until the line 7 passes through B. 
Then carefully prick through at the places where the lines 
between o and 7 cross AB. 

These ruled templates will be found very useful in graphical work, 
and it is worth while for the student to make them of celluloid, which 
may be done thus : Paper accurately ruled (or squared paper) may 
be readily obtained, or failing this, carefully set out ; the sheet celluloid 
is then placed over this and the lines scratched with a needle-point. 

Two sizes of templates should be made. The smaller one may have 
the twelve main lines 4" long and 0.2" apart, each division being sub- 
divided into five equal parts. Allowing a margin of h" all round, the 
size of the template would be 5" by 3V' outside. The larger template 
might conveniently have the twelve main lines 7" long and V' apart, 
each being subdivided into ten. The outside dimensions would be 8" 
by 5", leaving the same margin as before. 

The lines should be scratched of different widths for the sake of 
clearness. 



GENERAL INTRODUCTION 



ii 




Examples. 1. Draw a straight line, and on it mark off a length 
A B of 3.67". Bisect AB. 

Note. The bisection of a line is best done with the compasses, 
using the lead. Set the compasses to the half-length as nearly as 
can be guessed, and witli A and B as centres describe short arcs 
nearly meeting on the line. A second guess may now be made which 
is very nearly, if not quite, accurate, the arcs drawn, and the mid point 
between them estimated. 

2. Divide the line of Example I into 8 equal parts. 

Note. Here the method of continual bisection is useful. 

3. On a straight line mark off a length of 3.14", and divide this 

length into 5 equal parts, by each of the methods of this article. 



12 PRACTICAL PLANE GEOMETRY chap. 

8. Problem. To divide a line AB into consecutive 
parts which have given ratios to each other, say 2:3:5. 

(a) Draw a line through A inclined at any angle to AB, 
and to a convenient scale mark off A C, CD, DE, respect- 
ively equal to 2, 3, 5 units. 

Join EB and draw CT^and DG parallel to EB. Then 
AF-.FG: GB = AC : CD : DE = 2 : 3 : 5 (Euclid, VI. 10). 

(0) By the use of ruled tracing-paper or celluloid. This 
may be effected in a manner similar to that already de- 
scribed in Art. 7. The sum of 2 and 3 is 5, and that of 
2, 3, and 5 is 10. So by placing the line o over A, and 
the line 10 over B, the required points of division are 
obtained by pricking through where the lines 2 and 5 
intersect AB. 

If decimals occur in the given ratios, decimal subdivi- 
sions are required in the ruled template. 

9. Problem. The given line P represents a speed of 
24 6 feet per second ; construct the scale of speed, and 
measure the speed represented hy the line Q. 

(a) On any line set off RS equal to P, and draw ST 
perpendicular to RS. 

With centre R, radius 24.6 measured on a suitable 
scale, describe an arc intersecting ST in T, and join 
RT. 

Set the scale used in measuring 24.6 with its edge along 
RT, and mark or prick off the decimal divisions and sub- 
divisions shown. 

Perpendiculars from these to RS will give the required 
divisions of the scale of speed. The scale is then numbered 
as shown. 

Measuring Q on this scale, the speed represented by 
it is found to be 1 1.3 feet per second. 

(/>) By the use of a ruled template. Set the template 
with the zero line on R, and the 2.46 line over S, then 
prick through the decimal divisions and subdivisions of the 
scale. 



GENERAL INTRODUCTION 



13 




F G B 

Feet ptr second/ 




A 

1 



B 



r 



D 



Examples. 1. Divide a line 2. 87" long into two parts which shall 
be in the ratio 4:7. 

2. Divide a line 3. 19" long into two parts, so that the length of 

one part is to the length of the whole as 4 : 7. 

3. On a straight line mark off a length of 2.71", and divide the 

length into three consecutive segments, having the ratios to one 
another of 2.3, 3.6, 1.7. 

4. Measure the ratio of AC to AB in the bottom figure. Ans. 2.80. 

Place the zero line of the template on A, and rotate until one 
of the main division lines comes over B ; read off the position 
of C, and divide by B. 

5. Measure the ratios (a) AB-.AC : AD. Ans. 1 : 2.80 : 3.70. 

\b) DC : DB : DA. Ans. 1 : 3.06 : 4.20. 
(c) AC-.BD. Ans. 1.05 : 1. 



I4 PRACTICAL PLANE GEOMETRY chap. 

10. Problem. To construct a diagonal scale of 1J" to 
the foot, which shall measure feet, inches, and eighths of 
an inch. 

Set off eight equal divisions along a vertical line AB, 
and draw the nine horizontal lines through the points of 
division. 

Draw a series of vertical lines, \V' apart, for the main 
divisions of the scale, representing feet. 

Apply the |" scale to BC and AO, mark off twelve 
equal divisions on each of these lines, and draw the twelve 
diagonal lines between them. 

Figure the scale as shown. 

The length between the two dots represents i 2". 

11. Problem. Linear measure in Western India being 
as follows : 

1 tasu =1125 inches 

1 hath = 16 tasu = 1 foot 6 inches 

1 gaz =l l hath = 2 feet 3 ,, 

Draw a diagonal scale of -$ showing gaz, hath, and tasu. 

Mark off on it two lengths respectively equal to 2 gaz, 
hath, 12 tasu ; and 1 gaz, 1 hath, 4 tasu. (1898) 

Mark off sixteen equal divisions along a vertical line 
oZ>, and draw the seventeen equally spaced horizontal 
lines. 

By calculation, Ts V of 2' 3" = --" = 0.9 '. Therefore 
draw the vertical lines of the scale at horizontal intervals of 
0.9" to represent gaz. 

Set off DE= 1 hath = gaz = 0.6". Draw the diagonal 
line BO, and through 8 draw 81 parallel to BO. 

Figure the scale as shown. 

The two required distances are indicated on the scale 

by dots. 

Note. Diagonal scales have the disadvantage that the divisions do 
not extend to the edge, and they are not more accurate in use than 
ordinary scales. They are however useful where, as above, it is desired 
to represent quantities which are expressed in three denominations. 



GENERAL INTRODUCTION 



15 



B C 










3A 









/'l 


M 






f& 


M 






M 






4 


MM 






M 








M M M M M 







& 9" 6" 



tdAiV E 
16 

12 



O 

10 



/' 



D 



..... 1 

-X .,,,., -o - .. -., , 

N 1 



hcUhi 







zgaz 



11 



Examples. 1. Construct a diagonal scale, the representative 
fraction of which is -fa, reading yards, feet, and inches. 

2. Draw a scale showing hundredths of an inch by diagonal 

division. 

3. Draw a scale of feet to measure all distances between 70 feet and 

1 foot, where 5| feet is represented by .52 inch ; and by 
diagonal division, make this scale available for reading inches. 

4. Construct a scale of 120 feet to an inch, to measure 700 feet, 

from which single feet can be taken. 

5. Construct a scale of 100 fathoms, with iS fathoms represented 

by 1 inch, from which feet may be measured. 

6. Construct a scale of 76 miles to 1.3 inches, to read to single 

miles, and to exhibit 500 miles. 

7. A volume of 374 cubic inches is represented by a line 4.61 inches 

long. Construct a decimal scale of volume, and by its use 
measure the volume represented by a line 5.6 inches long. 
Am. 455 cubic inches. 



16 PRACTICAL PLANE GEOMETRY chap. 



12. Solution of the right-angled triangle. Many subse- 
quent problems reduce to that of solving a right-angled 
triangle, having given two of its elements. 

The jfo<? elements (exclusive of the right angle) are 



the height 


a, 


the base 


b, 


the hypothenuse 


c, 


the base angle 


A, 


the vertical angle 


B. 



There are nine possible cases. To solve the triangle 

i. Given the height and base a, b, 

2. given the base and hypothenuse b, e, 

3. given the hypothenuse and height c, a. 

4. Given the height and base angle a, A, 

5. given the base and base angle b, A, 

6. given the hypothenuse and base angle c, A. 

7. Given the height and vertical angle a, B, 

8. given the base and vertical angle b, B, 

9. given the hypothenuse and vertical angle c, B. 

The student will readily solve all these cases himself. 
We have placed them here for convenience of reference, 
and to direct attention to them. 

Examples. Solve the following four right-angled triangles by 
accurately drawn figures, and measure the results : 

1 Given the hypothenuse 3.2" and the height 1.9". 
Ans. b = 2.sf,A = z6.Ar\ 5 = 53-6. 

2. Given the height 2.18", base angle 36. 3 . 

Ans. = 2.97", 6 = 3.68", B = S3-7- 

3. Given the base 3.06", vertical angle 49.1. 

Ans. a = 2.66", c = 4.os", A =40.9". 

4. Given the hypothenuse 3.92", vertical angle 55.7. 

Ans, a -2.21", =3.24", ^ = 34.3. 



GENERAL INTRODUCTION 



17 




5. Measure all the sides and angles of the triangle ABC, the linear 

scale being j. Square the two sides a and b and add, and 
compare this with the square of c. 

6. The shadow cast by a vertical post 57" high on level ground was 

measured and found to be 87.6" ; find the altitude of the 
sun above the horizon. Ans. 33.1. 

7. Wishing to know the height of an electric arc light, I placed a 

5-feet rod vertically upwards on the floor, and found its shadow 
to measure 4.2 feet ; on moving the rod 6.2 feet along the 
floor directly away from the light its shadow measured 7.5 feet. 
Required the height of the light above the floor. Ans. 14.4 
feet. 

8. Two knots on a plumb-line at heights of 7 feet and 2 feet above 

the floor cast shadows at distances of 1 1.4 feet and 1. 65 feet re- 
spectively from the point where the line meets the floor. Find the 
height of the source of light above the floor. Ans. 12. 1 feet. 

9. A river AC, whose breadth is 217 feet, runs at the foot of a tower 

CB, which subtends an angle BAC of 27.4 at the edge of the 
opposite bank. Required the height of the tower. Ans. 112.4 
feet. 

10. A person on the top of a tower 68 feet high observes the angles 

of depression of two objects on the horizontal plane, which are 
in line with the tower, to be 32. 4 and 48. 6. Find their 
distance from each other and from the observer. Ans. 47.1 
feet, 90.7 feet, 127 feet. 

11. The hypothenuse of a right-angled triangle is 3.45 feet, and one 

of the sides is double the other ; determine the sides and angles. 
Ans. 1.54 feet, 3.08 feet, 26.5 , 63. 5 . 

12. The hypothenuse of a right-angled triangle is 43.5 feet, and 

one of the adjacent angles is double the other ; find the sides 
and angles. Ans. 21.7 feet ; 37-7 f ee t- 

C 



18 PRACTICAL PLANE GEOMETRY chap. 

13. Ways of defining angles. Let the student draw 
two straight lines at random, meeting at a point, and 
including some angle. By the use of any instrument 
except the protractor, let him obtain some information from 
the angle, which, being given to a neighbour, shall enable 
the latter to construct an angle of equal size. A number 
of different ways of doing this will now be given, but the 
student should think one or two out for himself before 
reading farther. 

Definitions. Let A OB be any angle. With centre O, 
and any radius, describe the arc RS, intersecting OB, OA, 
in R and S, and draw the chord RS. Then it will be 
quite clear that if the lengths of the radius OR and chord 
RS were given, this would be sufficient information to 
enable one to construct an angle equal to A OB. 

Again, in OB take any point P, and draw PN per- 
pendicular to OA, it is evident that a knowledge of the 
lengths of ON and NP would enable the angle to be repro- 
duced in size. 

We might, instead, give the lengths of NP and PO, or 
of ON and OP. Whatever pair we take, it will be a simple 
matter to construct an angle equal to A OB, remembering 
always that the angle ONP must be a right angle. 

But observe carefully that it is not the actual lengths of, 
say, ON and NP which are necessary, for the lengths of 
ON' and NP would do equally well. What, then, is it 
sufficient to give ? The answer is that we may give the 
ratio of ON to' NP, or NP to OP, or ON to OP, or RS to 
OS, or R'S' to O'S ; for then the student may take any 
convenient lengths for the pair of lines, so long as they are 
in the given ratio. 

For example, if ON were taken three inches and NP 
one inch, the same angle would be determined as by taking 
ON' 6 inches and N'P' 2 inches, 3 to 1 being the same 
ratio as 6 to 2. 

It will be convenient to give names to these ratios. The 
following are in constant use : 



GENERAL INTRODUCTION 



19 




S S'A 



RS 
OR 
NP 
OP 
ON 
OP 
NP 
ON 



is called the chord of the angle A OB 



sine 



cosine 



tangent 



These are abbreviated to cho AOB, sin AOB, cos AOB, 
and tan AOB respectively. Thus 



cho AOB = 
sin AOB = 



cos AOB = 
tan AOB = 



RS 

"OR 

PN 

hypothenuse OP 
base ON 



chord 

radius 

height 



hypothenuse OP 

height _ PN 

base ON 



Three-figure tables of the values of these ratios are given 
in the next article for angles between o and 90 , at intervals 
of one-tenth of a degree. 



20 



PRACTICAL PLANE GEOMETRY 



CHAP 



14. Trigonometrical Tables. 

SINES OF ANGLES 



0" 

10 
20 

30" 

40 
50 

60 

70 
80 





1 2' 3 


4 5 6 


T 8 9" 


Difference to be added. 


1 -2 -3 


4 -5 -6 


7 -8 -9 


.ooo 

.174 
342 

.500 

643 
.766 

.866 
94o 
.985 


.017 .035 .052 
.191 .208 .225 
358 -375 -39 1 
.515 .530 .545 
.656 .669 .682 
777 -788 -799 

.875 .883 .891 
.946 .951 .956 
.988 .990 .993 


.070 .087 .105 
.242 .259 .276 
407 -423 -438 

559 -574 -588 
.695 .707 .719 
.809 .819 .829 

.899 .906 .913 
.961 .966 .970 
995 -996 -99 8 


.122.139 -156 
.292 309 .326 
.454.469 .485 

.602.616 .629 

73W43 -755 
.839.848 .857 

.920.927 .934 
.974.978 .982 
.999 .999 1.000 


235 
235 
235 

1 3 4 
1 2 4 
123 

112 

Oil 

000 


7 910 
7 8 10 
689 

679 
5 6 7 
4 5 6 

3 4 4 
223 
111 


12 14 16 
12 13 15 
11 12 14 

10 11 13 
9 10 11 

7 8 9 

567 

3 4 4 
111 



COSINES OF ANGLES 








1 2 3 


4 5 6 


7" 8" 9 


Difference to be subtracted. \ 




V -2 -3 


4" -5 6 


V -8 9 



10 
20 

30 

40 
50 

60 
70 

80 


1.000 

985 
.940 

.866 
.766 
.643 

.500 
342 
174 


1.000 .999 -999 
.982 .978 .974 
934 -927 -920 
.857 .848 .839 

755 -743 -731 
.629 .616 .602 

485 -4 6 9 -454 
.326 .309 .292 
.156 .139 .122 


998 -996 -995 
.970 .966 .961 
.913 .906 .899 

.829 .819 .809 
.719 .707 .695 
.588.574 .559 

.438 .423 .407 
.276 .259 .242 
.105 .087 .070 


-993 -99o .988 
.956 .951 .946 
.891 .883 .875 

799 -788 .777 
.682 .669 .656 

545-530.5I5 

39i -375 -358 
.225 .208 .191 
.052 .035 .017 


000 

Oil 

112 

I 2 3 
I 2 4 

1 3 4 

2 3 5 
2 3 5 
2 3 5 


1 1 1 
223 

3 4 4 

4 5 6 
567 
679 

689 
7 8 10 
7 9 10 


111 

3 4 4 
567 

7 8 9 
9 10 11 

10 11 13 

11 12 14 

12 13 15 
12 14 16 



TANGENTS OF ANGLES 



0- 

10 

2D 


0" 


1 2 3 


4 5 6 


7 


8 


9 


1 
Difference to be added. | 


1 -2 -3 


4 -5 6 


7 -8 -9 


.COD 
.176 

34 


.017 .035 .052 
.194 .213 .231 
384 -44 -424 


.070 .087 .105 
.249 .268 .287 
.445 .466 .488 


123 
306 
.510 


.141 
325 
532 


.158 
344 
554 


245 
246 
246 


7 910 
7 911 
811 13 


12 14 16 j 

13 15 17 
15 17 19 


30 
40 
50 


577 
839 
i.ig 


.601 .625 .649 
.869 .900 .933 
1.23 1.28 1.33 


.675 .700 .727 
.966 1. coo 1.036 

I. 38 1.43 1.48 


754 
1.072 

i-54 


.781 
1. in 
1.60 


.810 
1.150 
1.66 


3 5 8 

4 7 11 
112 


10 13 16 

14 18 21 
233 


18 21 23 
25 28 32 

4 4 5 


60 


i-73 


1.80 1.88 1.96 


2.05 2.14 2.25 


2.36 


2.48 


2.61 








V0 J 
80 


2-75 
5-67 


2.90 3.08 3.27 
6.31 7.12 8.14 


3-49 3-73 4-oi 
9.51 11.43 !4-30 


4-33 
19.08 


4.70 
28.64 


5-i4 
57-29 






' 



GENERAL INTRODUCTION 



21 



CHORDS OF ANGLES 







1 






.ooo 1 


10 
20 


174 
347 


30 


.518 


40 
50 


.684 
845 


60 


1. 000 


70 
80 


1.147 
1.286 



Difference to be added. 



go 1 r 



9 !-l -2 -3 -4 -5 -6' -T -8 9 



.017 .035 .052 .070 .087 .105 
.191 .209 .226 .243 .261 .278 
.364 .382 .399 J .416 .433 .450 

534 -551 -568 j .585 .601 .618 
.700 .717 .733 j .749 .765 .781 
.8C1 .867 .892 : .908 .923 .939 

.015 1.030 1.045 I1.060 1.075 1.089 
.161 1. 175 1. 190 I1.203 1. 218 1. 231 
.299 1. 312 1.325 11.338 1. 351 1.364 



.122 
.296 
.467 

.635 
797 

954 



.140 

313 

.484 

.651 
.813 
.970 



'57 
33 
.501 

.667 
.829 
9S5 



1. 104 1. no 1. 133 
1.245 1.259 1.272 
1.377 1.389 1.402 



9 10 
9 10 
9 10 

8 10 
8 10 
8 9 

7 9 



12 14 ID 

12 14 16 
I2I4I5 

12 13 15 

II 13 14 

II 12 14 

IO 12 13 

IO II 12 

9 IO 12 



SINES, TANGENTS, CHORDS, AND RADIANS OF SMALL ANGLES 








1 


2 


3 


4 


5 


fi 


7= 


.go 


9 




0" 






















Diff. to be 
added. 





6' 


12' 


18' 


24' 


30' 


36' 


42' 


48' 


54' 


.0000 


.0017 


C035 


.0052 


.0070 


.0087 


.0105 


.0122 


.OI40 


oi57 


1' .0003 


1 


0175 


.0192 


.0209 


.0227 


.0244 


.0262 


.0279 


.0297 


.0314 


0332 


2' .0006 


2 


0349 


.0367 


.0384 


.0401 


.0419 


.0436 


.0454 


.0471 


.O489 


.0506 


3' .0009 


3" 


.052 


054 


.056 


.058 


059 


.061 


.063 


.065 


.066 


.068 


4' .0012 


4 


.070 


.072 


73 


75 


.077 


.079 


.080 


.082 


.084 


.085 


5' .0015 



Examples. 1. Required the sine of 34 5*. 

table of sines, we find 

Sine of 34 . .559 

Add for .5 diff. 7 



For the difference 1 1 add 
The required angle is 



3. Required the cosine of 414 J 



Referring to the 



Sine of 34.5 . .566 

2. Required the angle whose chord is -824, 

From the table, the chord of 48 is "813 

48.7- 



Cosine of 41 . . -755 

For excess .4 subtract diff. 5 
Cosine of 4 1. 4 . . .7 50 



22 PRACTICAL PLANE GEOMETRY chap. 

15. Problem. To set off any given angle, say 37 6. 

(a) By the protractor. If a circular protractor be used, 
then to ensure great accuracy, mark off two points for 3 7 "6, 
at opposite ends of a diameter. 

(b) By referetice to a table of sines. From the table the 
sine of 37.2 is seen to be .61 1. 

Now construct the right-angled triangle JVBO, making 
JVP=.6n,PO=i. Thus set off NP= 6. 1 1 andi 5 (9=io 
on the \" or |" scale. 

The angle PON equals 37. 6. 

(c) By reference to a table of cosines. From the table we 
find cos 37. 6 to be .793. 

To any convenient scale set off ON= .793 (or 7.93), 
and make OP=\ (or 10). 

The angle PONwW be 3 7. 6. 

(d) By reference to a table of tangents. We find from 
the table that tan 3 7. 6 = .7 70. 

Mark off ON= 1 (or 10), and JVP=.tjo (or 7.70); 
join OP. 

Then the angle PON= 37. 6. 

(e) By reference to a table of chords. The chord of 3 7. 6 
is found to be .645. 

With centre O, radius unity (or 10), describe the 
arc PP. 

With centre P, radius .645 (or 6.45), cut this arc in P ; 
join OP. 

Then angle POJV= 3 t.6. 

(/) By means of a scale of chords. A scale of chords 
marked CHO is generally given on a rectangular protractor. 

With centre 0, radius o to 60, describe the arc PP. 

With centre P, radius o to 37.6, cut this arc in P. 

Join OP, then the angle POJV= 37. 6. 

16. Problem. An angle AOB being given, to measure 
it in degrees. 

With centre 0, radius unity (or 10 read as 1) on any 
convenient scale, describe the arc PP. Draw the chord 
PE and the perpendiculars PJV, EQ. 



GENERAL INTRODUCTION 



23 




16 



(a) Measure the angle directly with the protractor 

(b) or measure PN and refer to the table of sines ; 
if) or measure ON and refer to the table of cosines ; 
if) or measure QE and refer to the table of tangents ; 
(e) or measure PE and refer to the table of chords ; 
if) or measure PE on the scale of chords, the arc PE 

having been struck with the radius o to 60. 

Examples. 1. Draw an isosceles triangle with sides of 5", 5", 
and 2.2". Determine and measure the vertical angle by each 
of the six methods of this article. Take the mean of these, 
and observe by how much each separate result differs from the 
mean. A/is. The angle is 25. 4 . 

2. The sine of an angle is .820, what: is the tangent? Aris. 1.44. 

Note. This example should be worked in two ways : first, by 
construction ; secondly, by an inspection of the tables. 



24 PRACTICAL PLANE GEOMETRY chap, 



17. Miscellaneous Examples. 

The student, from his previous study, should be able to work such 
examples as the following : 

1. Determine by construction a perpendicular through a given 

point to a given line, for various positions of the point and 
restrictions as to the length of the line. 

2. Through a given point draw a line to meet a given line at a 

given angle. 

3. Draw a triangle, having given {a) the three sides ; (/') two sides 

and the included angle ; (c) two sides and an angle opposite to 
one ; (</) two angles and the intermediate side ; (<") two angles 
and a side opposite to one of them. 

4. Construct an irregular quadrilateral, having given (a) the four 

sides and one diagonal ; (/') the four sides and one angle ; (r) 
three sides and the two diagonals ; (J) three sides, one diagonal 
and one angle ; (e) three sides and two angles, etc., etc. 

5. Construct an irregular pentagon, having given the five sides and 

two diagonals ; the five sides and two angles, etc. 

6. Construct any irregular polygon, having given adequate data as 

to the sides, diagonals, and angles. 

7. ABCDE are the five consecutive corners of an irregular pentagon. 

Construct the figure, having given 

Sides- .// = 2.qi"; BC=2.Sl"; C/^3.56"; ^^=1.34". 

Diagonals AC = 4. 63"; ^ = 3.98". Angle DEA = 133.2. 

The polygon is not to have any internal angles greater than 
180 (called re-entrant angles). 

Measure the remaining side and diagonals. 
Am. EA = 3.04"; CE = 4. 57; J) A = 3. 66". 

8. Set out all the pentagons which comply with the data of Ex. 7, 

re-entrant angles being allowed. 
Note. There are four different polygons with re-entrant angles. 

9. Copy a given triangle so that a specified side shall have an 

assigned position. 

10. Copy a given quadrilateral so that a specified side or diagonal 

shall occupy an assigned position. 

11. Copy any given rectilinear figure so that a specified line shall 

have a given position on the paper. 

12. Draw a triangle similar to a given triangle, the length and 

position of one side of the former being given. 

13. Copy the figures on page 9, (a) the same size ; {!>) double size. 

14. Copy the figures on page 27, (a) the same size ; (/>) double size. 

15. Reduce any given quadrilateral to a triangle of equal area ; and 

reduce the triangle to an equivalent rectangle. 

16. Find the centre of any given circle or circular arc. 



GENERAL INTRODUCTION 25 



17. Describe a circle 4^" diameter. Suppose this circle to be 
given to you on the paper, and that you have only a pencil 
and a ruler with two parallel edges 1 j" apart ; show how the 
centre of the circle can be obtained. (1888) 

18. Determine a circle to pass through three given points, or draw 
the circle which circumscribes a given triangle. 

19. Draw a circle to touch three given lines, or determine the 
inscribed and escribed circles for a given triangle. 

20. Draw a tangent to a given circle from a given point on the 
circumference. 

21. Find the point of contact of a tangent to a circle drawn from 
an external point. 

22. Draw a circle of given radius to touch (a) a given line at a 
given point ; (6) a given circle at a given point. 

23. Draw a circle of given radius {a) to pass through two given 
points ; (6) to touch two given lines ; (c) to pass through a 
given point and to touch a given line. 

24. Draw a circle of given radius (a) to touch two given circles ; 
(6) to touch a given line and circle ; (c) to pass through a given 
point and touch a given circle. 

25. 1 >raw a circle of given radius which shall have its centre on a 
given line, and touch (a) a second given line ; (/') a circle. 

26. Illustrate the theorem of Note 1, Prob. 21, in the following way : 
Draw on tracing-paper two lines AP, AQ meeting at any 
angle, say make A = 30, and AP, AQ, 3" and 4". Between 
AP and AQ draw a series of lines P t Q v 7>. 2 Q.,, P 3 Q 3 , . . . 
parallel to J'Q. These may be about J" apart" Glue a small 
piece of paper on the tracing at A for strength. 

Next draw in ink on paper any figure, say a semicircle 3" 
diameter. Mark a point O inside the semicircle, say near the 
middle ; place the tracing with A at 0, and insert a pin at 
this point. 

Then rotate the tracing, and as points on one of the lines, 
say on AP, come in succession on the boundary of the figure, 
prick through the corresponding points on the other line AQ. 
The locus of the latter will be a semicircle, larger in the linear 
ratio 3 : 4, and turned through an angle A about O. 

Again pin the vertex A of the tracing to a point outside the 
semicircle, say near one end. Rotate, and as the points Q 
come on the boundary, prick through the corresponding points 
P. A copy of the semicircle will again be obtained ; this time 
reduced in the ratio 4 : 3, and turned through an angle A. 



CHAPTER II 



SIMILAR RECTILINEAL FIGURES AREAS 

18. Similar polygons. Definition. Two equiangular 
polygons which have the sides about their equal angles 
proportional are said to be similar. 

Thus the pentagons ABCDE and abcde are similar, being equi- 
angular and having 

AB :BC :CD : DE :EA=ab: be : cd : de : ea ; 
AB _BC _CD _DE EA 

ab be cd de ea ' 

Except in the case of the triangle, polygons may be equiangular 
without being similar, for example, ABCDE and A'B'C'D'E'. And 
evidently, with the same reservation, the sides may be proportional 
without the corresponding angles being equal. 

We now give some of the more important properties of similar 
figures, on which the constructions of many problems are based. 

An instrument called a pantograph, for the tracing of similar figures, 
is sometimes used. 

Theorem i. If two lines be cut by a series of parallel lines, 
the ratios of corresponding segments are all equal. 

_ KL LM MN KN LN OK OL 

thus ,,=, = = . =f = = = . . . etc. 
// I in mil kn hi ok ol 

Theorem 2. If two polygons have their sides respectively 
parallel, and the lines joining their corresponding corners all 
converging to a point, the figures are similar ; and conversely. 



ch. ii SIMILAR RECTILINEAL FIGURES AREAS 




w 


In 


\ 


m\ 


im 


L \ 


I 




Thus the polygons PQRS and pqrs which satisfy this condition are 
similar, and the following relations hold : 

PQ_ Q_#s_sp_qp_ OQ _ OR _ OS 

pq qr rs sp Op Oq~ 0r~~ Os' 

The point O is called the pole. 

The student will find many illustrations of this theorem. Thus in 
the figure of Art. 6 the pole is situated within the polygon. In 
Prob. 36 the pole coincides with one corner A. In Prob. 24 the 
pole is inaccessible ; it might be at an infinite distance away, when the 
lines through it would be parallel. 

Theorem 3. The areas of similar figures are propor- 
tional to the squares on any two corresponding sides. 



Thus 



area PQRS 
area pqrs 



{PQf 



KQK) 



{qr? 



{OPf 
(Op?' 



etc. 



or 



area ONn _ (JVn) 2 _ {ON? 

area OKk ~JM?~{ORy i= CtC " 



28 PRACTICAL PLANE GEOMETRY chap. 

19. Problem. Two triangles ABC, DEF are given. 
It is required to draw a triangle def with its vertices 
d, e, f in BC, CA, AB, and its sides de, ef, fd parallel to 
DE, EF, FD. 

Between AB and AC draw any line F ' E' parallel to 
FE ; draw FD, E ' F>' parallel to FD, ED. 

Join AD', and let this line (produced) meet BC in d. 
Through d draw de, ^/"parallel to DE, EF, and join ef. 
Then def is the triangle required. 

20. Problem. A triangle ABC and a quadrilateral 
DEFG are given. It is required to draw a quadrilateral 
defg similar to DEFG, with its side de in AB, and its 
vertices f, g in BC, CA respectively. 

Copy the figure DEFG in the position D'E'FG' ; that 
is, with D'E' in AB, and G' in AC. 
Draw AF' to intersect BC inf. 
Then draw/?-, gd, fe parallel to F G' , G'D', FE'. 
We thus obtain the inscribed quadrilateral as required. 

21. Problem. Two triangles ABC, DEF are given. 
It is required to draw a triangle def similar to DEF, with 
f at a given point in AB, and d, e in BC, CA respectively. 

From the given point / draw fE' to meet BC at an 
angle fE B FED. From E' draw E'e, making the angle 
CE'e = DFE. 

Join ef. From e draw ed, where the angle fed = FED. 
Join df. Then def is the triangle required. 

Note I. This solution is based on the following theorem. 

Theorem. If a triangle of varying size but constant shape rotate or 
swing about one vertex A, while a second vertex P traces any figure, 
then the third vertex Q traces a similar figure turned through an angle 
A, the linear dimensions of the two figures being in the ratio AP : AQ. 
The figure shows two positions fD'E, fde of such a triangle. As 
one vertex moves from D' to d, the other moves along E'e, inclined 
at /3 to D'd. See also Ex. 26, p. 25. 

Note 2. With modified data some of the points would be situated 
in the sides AB, BC, CA produced. The student will now be able to 
work examples relating to triangles, squares, parallelograms, etc., in- 
scribed in and circumscribed about triangles. 



ii SIMILAR RECTILINEAL FIGURES AREAS 



29 




Examples. 1. Draw a 



BC= 3 .S", CA = 2.$' 



triangle ABC, making AB = 4", 
In ABC inscribe an equilateral 



triangle with one side inclined at 45 to AB. 
In the triangle of Ex. 1 inscribe a square with one side in AB. 
In ABC inscribe an equilateral triangle with one vertex bisecting 

AB. 
In ABC of Ex. 1 inscribe a similar triangle abc, with c at the 

middle point of BC, a in AB, and b in AC. 
In ABC inscribe a parallelogram with sides in the ratio 2:3, and 

included angle 60, a longer side lying in BC. 
Draw a circular arc, 3" radius, and two radii including 6o. In 

this sector inscribe a square with a side along a radius. 



30 PRACTICAL PLANE GEOMETRY chai>. 

22. Problem. Two lines AB and CD and a point P 
are given. It is required to draw through P a line 
terminated by the given lines, and divided at P into two 
segments which are in a given ratio, say 2 : 3. 

Through P draw any line meeting one of the given 
lines, say AB, at E. 

Bisect PE at G, and set off PE= 3 PG. Draw ES 
parallel to AB, meeting CD in S. 

Then the line through 6" and P meeting AB in R is 
the one required. 

For, since the triangles PES, PER are similar, 
PR _PE _2PG _2 

P~S~PF~'^PG~3 

Note. If PF were made equal to PE, then SR would be bisected 
at P. 

23. Problem. To draw the straight line bisecting the 
angle between two given straight lines AB and CD, the 
intersection of the latter being inaccessible. 

Draw any line EF between the given lines, and bisect 
the angles AEE and CFE by lines EO and FO meeting 
at O. 

Also bisect the angles BEE and DEE by lines EO' 
and EO' meeting at O'. 

Then the line 00' will be the line required. 

For it is obvious that O is equidistant from AB, EF, and CD ; 
so also is the point O'. 

Hence 00' bisects the angle between AB and CD. 

24. Problem. Having given two straight lines AB 
and CD and a point P, it is required to draw the straight 
line through P, such that the three lines converge to the 
same point, the point being inaccessible. 

Take any convenient points E and E in AB and CD 
respectively, and join EF, FP, and PE. 

Parallel to EF draw any corresponding line GH; draw 
GQ, HQ parallel to EP, FP. Join PQ. 

Then the lines AB, CD, and PQ if produced would all 
meet at the same point. 



II 



SIMILAR RECTILINEAL FIGURES AREAS 




P\ 



Q 




H B 



Examples. 1. Draw a quadrilateral ABCD of the following 
dimensions : 

Sides AB=i.f; BC=t,.o"; CD =3.40"; ^ = 3.30". 
Diagonal BD = ^.^.". Draw the two bisectors of the angles 
between AB, DC, and AD, BC. 

2. Let the diagonals of ABCD, Ex. I, intersect in P. Through 

P draw the two lines which converge respectively to the same 
points as the pairs of opposite sides. 

3. Through the point P, Ex. 2, draw the two lines between the 

pairs of opposite sides of the quadrilateral which are bisected 
at P. 

4. Draw an equilateral triangle ABC of 3" side, and take points D, 

E in AB, AC distant 1.9" and 1.8" from A. Determine the line 
through A which converges to the same point as BC and DE. 



32 PRACTICAL PLANE GEOMETRY chap. 

25. Problem. To find a fourth proportional D, to three 
given lines A, B, and C ; that is, to find a line D such 
that A : B : : C : D. 

Take any two lines intersecting at a point O ; along one 
set off OA, OC equal to A and C respectively ; and along 
the other set off OB equal to B. Join AB, and draw 
CD parallel to AB. 

Then OD is the required length of D. 

If AE be drawn parallel to CB, then OE is the length 
of a line E such that C : B : : A : E. 

26. Problem. To find a third proportional C, to two 
given lines A and B ; that is, to find a line C such that 
A : B : : B : C. 

Take any two lines intersecting at O ; along one set off 
OA, OB' equal to A and B respectively ; and along the 
other set off OB equal to B. 

Join AB and draw B'C parallel to AB. Then OC is 
the required length of C, such that A : B : : B : C. 

27. Problem. To find a mean proportional C between 
two given lines A and B ; that is, to find a line C such 
that A : C : : C : B. 

Take a point O in any line, and set off in opposite 
directions OA equal to A, and OB equal to B. 

Describe a semicircle on AB. Then OC, drawn perpen- 
dicular to BA, is the required length of C (Euc. VI. 13). 

C is also called the geometrical mean between A and B. 

28. Problem. To divide a line AB in extreme and 
mean ratio ; that is, to find a point C in AB such that 
AB : AC : : AC : CB. 

Bisect AB at D ; draw BE perpendicular to AB, and 
equal to BD, and join EA. With centre E describe arc 
BE; and with centre A describe arc EC. 

This gives C, the required point of division (Euc. 
VI. 30). 



II 



SIMILAR RECTILINEAL FIGURES AREAS 



33 



By 

A i 







c / 

A\ 



V 







E B 



D 



A / \ 



\ \ 



26 

P 



Q 



R 



25 



B C 



A 



B V 



B 



-Bi 









n 




s 






s 




% 


















/ 






\ 


/ 
/ 






\ 


1 






\ 


1 






\ 


I 

t 




\ 


1 

I 
J 



27 



AT 



\ \ 



-< i- 



A 

E 



D C 28 ^ 



Examples. 1. Find the fourth proportionals to P, Q, R ; to 
A, Q, A; and to Q, P, R. Ans. .92"; 2.7S" ; 1.45". 

2. Find the third proportionals to P and Q ; to Q and A; and to 

A' and P. Ans. 1.28"; 2.53"; 3.48". 

3. Find the mean proportionals between P and (2 ; Q and A' ; and 

A and P. Ans. 1.79"; 1.36"; 1.52". 

4. Divide A in extreme and mean ratio, and give the value of the 

ratio. Ans. 1.62. 

D 



34 PRACTICAL PLANE GEOMETRY chap. 

29. Problem. To find the harmonic mean between 
two given lengths a and b ; that is, to determine a length 
k such that b-k:k-a::b:a. 

From any point H in a straight line mark off ffA, HB 
equal to a and b. 

Through A and B draw any two lines AO, BO per- 
pendicular to one another. Join OH and set off the angle 
A K equal to the angle A Off. 

Then HK is the required length of k, the harmonic 
mean between a and b ; and ffA, HK, HB are in 
harmonic progression. 

.Votes. The relation b 'k ~. k a = b : a is equivalent to 
KB _HB _ AH _BH . 
AK~ HA~' ,OX J^A~ ~BK ' 

and comparing these equations we see that BA', BA, BH are also in 
harmonic progression. 

The line AB is said to be harmonically divided in H and A~ and 
the line HK in A and B. And four such points are spoken of as a 
harmonic range. 

The series of lines or pencil of rays formed by joining II, A, K, B 
with any point 0' is called a harmonic pencil, because any transverse 
line or transversal can be shown to be cut harmonically by the pencil. 

Thus H ', A', A"', B" and B x , II 1 , ./ x , A\ are harmonic ranges. 
Stated formally 

Theorem I. A harmonic pencil divides all transversals har- 
monically. 

The alternate points or rays are said to be conjugate to one another. 
Thus A and B, or 0.1 and OB are conjugate. Likewise OH and 0A~. 

Theorem 2. A transversal which is parallel to a ray is bisected 
by the conjugate. 

Thus the transversal A'^H^ is parallel to O'B, and is bisected in 
A . We also have 

Theorem 3. If a transversal parallel to any one of the four rays 
of a pencil is bisected, the pencil is harmonic. 

Thus if OC bisect any angle A OB, and OD be taken perpendicular 
to OC, the pencil is harmonic because any transversal parallel to OD 
or OC is bisected. 

A cotnplete quadrilateral affords an example illustrating harmonic 
pencils. See the figure. 

Let the diagonals of any quadrilateral PQRS intersect in N, and 
the pairs of opposite sides produced in Z, M. Join 71, J/, N. Then 
the pencils radiating from L, M, N are all harmonic. 



II 



SIMILAR RECTILINEAL FIGURES AREAS 



35 




Examples. 1. Take a and b each double the length given in the 
figure above, and find the harmonic mean k. 

2. Find the arithmetical mean r, and the geometrical mean e; 

between a and b, and vertify the theorem that /-, g, and / are in 
geometrical progression. 

3. Draw any complete quadrilateral PQRSLMlV, and test whether 

the pencils through L, M, A^are harmonic, by applying Theorem 
3 of this article. 

4. Taking the lines a and b in the figure above, find a line c such 

that a, b, c are in harmonic progression. Ans. - 2.46". 

5. Draw OA, OH, OJ3, making the consecutive angles A OH, 

HOB equal to 25 and 55 respectively. Find the fourth ray 
of the harmonic pencil, and measure the angle it makes with 
OB. Ans. 70. 8. 

6. Through the point b, Fig. 6, p. 47, draw a straight line cutting oa, 

oc in d and e, such that the triangles odb, obe are equal in area. 

7. Draw a line AB 2" long, and divide it internally and externally 

at K and H into segments which have the same ratio, say 3 : 4. 



36 PRACTICAL PLANE GEOMETRY chap. 

30. Problem. To reduce a given polygon ABCDEF 
to a triangle of equal area. 

Join D to B, A, and F Draw CG parallel to DB to 
meet AB produced in G, and join DG. Draw EH parallel 
to DF to meet AF produced in H, and draw HK parallel 
to DA to meet BA produced in K. Join DK. 

Then DKG is a triangle having an area equal to the 
polygon. 

This solution is based on the theorem that triangles on the same 
base and between the same parallels are equal in area. 

31. Problem. To construct a square equal in area to 
a given rectangle. 

Determine a mean proportional between the sides of 
the rectangle (Prob. 27); this is equal to the side of the 
required square. 

32. Problem. To construct a square equal in area to 
a given triangle. 

Let a rectangle be drawn, one side of which is equal to 
the base of the given triangle, and an adjacent side equal 
to half the altitude. 

The rectangle so drawn is equal in area to the triangle. 

Hence the problem reduces to the last one. 

33. Problem. To construct a square equal in area to 
a given polygon. 

Reduce the given polygon to an equivalent triangle 
(Prob. 30). Then apply Problem 32. 

34. Problem. To construct a rectangle equal in area 
to a given rectangle, and having one side equal to a 
given line. 

Let ABCD be the given rectangle, and AF the given 
side of the required rectangle. 

Draw FF equal and parallel to BC ; join FA cutting 
BC at H; through //draw MK parallel to AF. 

Then ME is the required rectangle. 



II 



SIMILAR RECTILINEAL FIGURES AREAS 



37 



L -" 


,,'-'' 


// 



34 B 




35. Problem. To divide a given polygon ABODE into 
a number of equal areas by lines drawn from a given point 
P in one side. Say into seven parts. 

As in Prob. 30, draw EF parallel to PA ; and DG, 
GH parallel to PC, PB, thus reducing the polygon to an 
equivalent triangle with vertex P and base FH. 

Divide FH into the required number of equal parts. 

Then, reversing the construction, draw n' parallel to 
FE ; 44', 5 5', 66' parallel to HG ; and 6' 6" parallel to GD. 

Join P to i', 2, 3, 4', 5', 6", and the problem is solved. 

Examples. 1. Draw a pentagon ABCDE as follows : 
Sides AB=i$'; BC=2" ; CD = 2\" ; DE=i\". 
Angles ABC=\2o; B CD = 8o ; CDE= 125." 
Reduce the figure to an equal triangle with vertex D and 
base along AB. Then reduce the triangle to an equal square. 
2. Divide the pentagon of Ex. 1 into eight equal parts by lines 
drawn through the middle point of DC. 



38 PRACTICAL PLANE GEOMETRY chap. 

36. Problem. It is required to draw a polygon 
similar to a given polygon, and having an area which 
bears to the area of the latter a given ratio, say 3 : 5. 

Let ABCDE be the given polygon. In any side BA, 
produced in either direction, say beyond A, set off Ap and 
AP equal to 3 and 5 on any suitable scale. 

On Bp, BP draw semicircles as shown ; and from A draw 
a line perpendicular to AB, meeting these semicircles in q 
and Q. Join QB, and draw qb parallel to QB. 

Then Ab is one side of the required figure, and the 
latter AbcdeA is completed by drawing the diagonals AC, 
AD, and then the sides be, cd, de, parallel to BC, CD, DE. 

For (Aqf = AB.Ap, (Euc. III. 35, or Prob. 27), 

and (AQ? = AB.AP; 

area AbcdeA {Abf {Aqf AB-Ap _Ap _3 



then 



area ABCDEA {AB) 2 {AQf AB-AP AP 5 



37. Problem. To divide a given triangle ABC into 
two equal areas, by a line drawn through any given 
point P. 

Bisect the sides of the triangle in D, E, F. 

Join AD, BE, CF and bisect these lines in G, PP, K. 

Join HK, intersecting OD in L, and bisect OL in AP. 
And through H, M, K draw a fair curve touching BH and 
CK2X //and K. 

In like manner draw the curves GH, GK. 

Through P draw PQ so as to touch one or other of 
the three curves, according to the situation of P. 

Then PQ divides the triangle into two equal areas. 

A T ote. The true curves should be hyperbolas, to which the sides of 
the triangle are asymptotic (see Chapter IV.). The bisection of OL in 
M makes them parabolas, from which the true curves for such small 
arcs do not visibly differ. 

Alternative Solution. Along AB, AC set off AN, AN each equal 
to the geometrical mean between AB and AE. Bisect NN in T. Join 
A T and produce to S, making AS=AN Then T is the vertex, S a 
focus, and A the centre of the hyperbola KH, and the tangent through 
P may be found by a construction analogous to that of Prob. 94. 



SIMILAR RECTILINEAL FIGURES AREAS 



39 




Examples. 1. Draw a triangle ABC having given AB^.o", 
BC=t,.2", CA= 1.8". Divide the triangle into two equal 
areas by a line parallel to AB. 

2. Divide the triangle of Ex. i into five equal areas by lines parallel 

to BC. 
Note. Divide AB into five equal parts, and from the points of 
division draw perpendiculars to meet a semicircle on AB in 
I, 2, 3, 4. With centre A draw arcs through I, 2, 3, 4 to 
meet AB in 1', 2', 3', 4'. Draw lines through the latter points 
parallel to BC. 

3. Draw a pentagon ABCDE as follows : 

Sides AB= 1. 7" ; BC=i.3$" ; CD= 1.45"; >= 1.65" ; 
EA=.$". Diagonals AC= 2.5", AD=2.2". 

Draw two similar pentagons, one 2 and the other 4- the area 
of ABCDE. 

4. Divide the pentagon of Ex. 3 into four equal areas by lines 

parallel to the sides. 

5. Divide the triangle of Ex. 1 into two parts of equal area, by a line 

passing through a point /"outside the triangle, the position of 
/being given by AP= 1" ; BP= 2.8". 



4 o PRACTICAL PLANE GEOMETRY chap. 

38. Problem. To find the number of square units of 
area in any given polygon. 

Let ABCD be the square equal in area to the polygon, 
obtained as in Prob. 2>e>- 

Produce one side DA to *S", making AS one unit of 
length. Join SB, and draw BT perpendicular to BS. 

Then AT, measured on the scale having AS as unity, 
gives the number of square units of area required. 

Thus if AS be set off i", the area in square inches is obtained by 
measuring A T on the i" scale. Or if AS be I centimetre long, then 
AT measured on the centimetre scale gives the area in square centi- 
metres. 

Note that A T is a third proportional to AS and AB ; and that AB 
is a mean proportional between AS and AT. 

To construct a square of given area, say 3 square inches, draw a 
rectangle 3" long and 1" broad, and reduce to an equivalent square. 
Or set off AT= 3", AS= 1", and find the mean proportional AB. 

39. Problem. To construct a polygon similar to a 
given polygon ABCD, and having an area equal to that 
of a given polygon Q. 

As in Prob. 40, determine AF, the side of a square equal 
in area to Q; and also AG, the side of a square equal in 
area to ABCD. (Construction not here shown.) 

Set off AB and AG along a line AE, making any angle 
with AB ; join GB, and draw Fb parallel to GB. 

Join AC, and draw be and cd respectively parallel to BC 
and CD. Then Abed is the required rectilinear figure. 

area of Abed Al> 2 AF 2 area of Q 



For 



area, of ABCD AB 2 AG 2 area of A BCD' 



It is important that the reader should observe that this problem is 
the general case of a distinct type or class of problems, and that the 
method of solution is of a general character. Two figures, which may 
vary in size and shape, are given, and a third figure has to be deter- 
mined, such that it is similar to one of the given figures and equal 
in area to the other. This is the type of problem. The data of 
the problem may be varied considerably, in so far that two triangles, 
rectangles, squares, etc., or any combination of these, may be substi- 
tuted for the pair of figures given above ; but the problem remains 
cf the same character, and the same method of solution is employed. 



ii SIMILAR RECTILINEAL FIGURES AREAS 41 



D 39 




Such observations should render it unnecessary to explain in detail 
the solutions of further problems of the same kind, such as the 
examples below. 

It may here be remarked that the student should acquire the 
habit of closely observing the type of any problem which comes under 
his consideration, and constantly aim at classifying problems, with the 
object of detecting general methods where such may exist, rather than 
to regard each problem as having no relation or resemblance to 
previous problems. In this manner the power of successfully attacking 
examples may often be greatly increased. 

Examples. 1. Construct an isosceles right-angled triangle equal 
in area to a given regular pentagon 1 j" side. 

2. Construct an isosceles triangle, vertical angle 40 , equal in area 

to a triangle whose sides are 2^", i||", and 2. 1". 

3. Construct a rhombus, one angle of which is 6o, equal in area 

to a given rectilineal figure ABCD, of which AD is 2", angle 
BAD 105 , AB 2", BC 3I", DC 1.7". 

4. Determine an equilateral triangle having an area of 4 square 

inches. Measure the side. Ans. 3.04". 

5. Construct a regular heptagon having an area of 5 square inches. 

6. Construct a rectangle equal in area to a hexagon of It" side, 

the ratio of the sides of the rectangle being 2 : 3. 

7. Determine graphically the area of the quadrilateral ABCD of 

Ex. 3, in square inches and also in square centimetres. Ans. 
4.62 ; 29.8. 



42 PRACTICAL PLANE GEOMETRY chap. 

40. Problem. Having given two similar rectilinear 
figures, it is required to construct a similar figure, the 
area of which shall be equal to the sum or difference of 
the areas of the given figures. 

Let ABCDE, Abcde be the two given similar figures. 

For the sum draw BN at right angles to AB, and 
make BN= Ab. 

With A as centre, describe the arc NB' to meet AB 
produced in B' ; draw B'C, CD', D'E' respectively parallel 
to BC, CD, and DE, to meet AC, AD, AE produced in 
C ' , D ' , and E ' . Then A'B'CD'E' is the required figure. 

For the difference describe a semicircle on AB as 
diameter. 

With centre B, radius Ab, cut this semicircle in M; 

With centre A, radius AM, draw the arc MB". 

Then a polygon with AB" as one side, and similar to 
the others, is the figure required. 

Note. These constructions are based on the following theorem, of 
which Euc. I. 47 is a particular case. 

Theorem. If similar polygons be described on the three sides of a 
right-angled triangle ABC, so that AB, BC, CA are corresponding 
sides of the three figures respectively, then the areas of the polygons 
described on the sides AB and BC are together equal to the area of that 
described on the hypothe?iuse AC. 

41. Problem. Having given any three rectilinear 
figures, which may be denoted by A, B, and C, to deter- 
mine a fourth figure which shall be similar to C, and 
have an area equal to the sum or difference of the areas of 
A and B. 

Determine, by means of Prob. 39, a figure E similar to 
A, and equal in area to B. 

By Prob. 40 determine a figure F similar to A (or E), 
and having an area equal to the sum or difference of the 
areas of A and E. 

Finally, by means of Prob. 39, determine a figure G 
similar to C, and having an area equal to that of F; then 
G is the required figure. 



ii SIMILAR RECTILINEAL FIGURES AREAS 43 




G E. 



C! A 



B 

\h 




42. Problem. To find a rectangle equal in area to a 
given trapezoid ABCD. 

Let AD, BC be the parallel sides of the trapezoid. 

Bisect AB, CD in E, F. Then EF is parallel to AD 
and BC, and is midway between them. Through E and 
F draw GH and KL perpendicular to EF. 

Then the rectangle GL is equal in area to the trapezoid, 

and the area is equal to EFx KL, or - x KL. 

2 

Examples. 1. Draw two equilateral triangles respectively equal 
to the sum and difference of two equilateral triangles of 2" and 
3" sides. 

2. Draw two circles respectively equal to the sum and difference 

of two circles of lA" and 2|" diameters. 

3. Draw a square equal to the sum of the areas of two equilateral 

triangles of 1.7" and 2.8" sides. 

4. Draw an equilateral triangle equal to the difference of two 

squares of 1.2" and 2.7" sides. 

5. Draw a square equal to the sum of an equilateral triangle and 

a regular pentagon, each of ii" side. 

6. Draw an equilateral triangle equal to the difference of a square 

and a regular hexagon, each of 2" side. 



44 PRACTICAL PLANE GEOMETRY chap. 

43. Problem. To determine approximately the area 
enclosed by a given irregular curve AB, a base line CD, 
and two perpendicular ordinates CA, DB. 

A perpendicular erected at any point on the base CD of the figure 
is called an ordinate. 

Draw a series of equidistant ordinates between CA and 
DB, dividing the area into a number of strips of equal 
width ; in this case six. One of these strips is shown with 
shade lines. 

Draw a second series of intermediate ordinates, midway 
between those first drawn, and let y v y 9 , . . . y 6 denote 
their lengths. 



Determine the mean ordinate y, 



_ Jl+J'-2 + - -+}\ 



6 



6 

Take CK equal to y m and through K draw KL parallel 
to CD. Then the area of the rectangle KD is equal to 
area required. 

The result is only approximate, because the upper boundary lines of 
the strips are curved instead of straight. By increasing the number of 
strips, and so making the width of each less, the error can be reduced 
to a very small proportion of the total area. 

We now state some other rules, available for calculating approxi- 
mately the mean ordinate y m ; that is, the height CK of the equivalent 
rectangle. 

In employing these we require the ordinates which bound the strips 
(not the middle ordinates of the spaces). In the figure there are 
seven such ordinates to the six divisions, CA being the first, and DB 
the last. These may be denoted by A , /i v h. h % , //.,, // 5 , h 6 . 

If there were n equal divisions, or strips, there would be ;/ + I 
ordinates 

"o> "i> "2> " 1> " 

With this notation, the rules in question maybe written as follows : 
Ordinary rule. Any number of divisions. 

?m = I (i U'O + h n) + h +*, + *,+ . . . + V-l} 

Simpson's Ji ist rule. Number of divisions even. 
When n = 2. y m = \ (// + 4 /i l + h 2 ). 

When n~2, or 4, or 6, etc. 

?>* = {b + A n +2 (^ 2 + ^4 + - +*-) + 4(* 1 + A 8 + . -+K-l)}- 



ii SIMILAR RECTILINEAL FIGURES AREAS 45 




Simpsotis second rule. Number of divisions a multiple of 3. 

When n 

When =3, or 6, or 9, or 12, etc. 



J'm = k ( 7/ + Z h \ + 3 A 2 + * 3 ). 



+ K-z) + 3(^1 + * a + ; h + h + - + K-*. + K-i)} 

Weddle's rule. Number of divisions a multiple of 6. 

When n = 6. y m = fa {A Q + K + A t + h Q + 5 (A + h z + h b ) + h 3 } . 

When h = 6, or 12, or 18, etc. 

J ' = Tcb^ + / ' 2 + - +*+5(*i+*, + . 

+ Vl)+^3 + 7 '6 + - -+K-Ji- 

Examples. 1. Draw a quadrant of a circle 3" radius. Divide 
one of the radii into six equal parts, and draw the ordinates, 
which will be parallel to the other radius. Determine the 
mean ordinate by each of the rules given, and compare the 
results. Set out the equivalent rectangle. Reduce to an 
equivalent square. And find the area of the quadrant in square 
inches. Ans. Mean ordinate = 2. 36". Side of square = 2. 66". 
Area =7. 07 sq. inches. 

2. Plot the curve for which thirteen equidistant ordinates, spaced 

i" apart, have the following successive values : 

".4.1", 4-o", 4.0", 3-9"; 3-55", 3-16", 2.73", 2.34", 2.05", 

1.77", i.6o", 1.45", 1.0". 

Obtain the mean ordinate by each of the rules given. Draw 
the equivalent rectangle, and determine the area under the 
curve in square inches. Ans. Mean ordinate = 2. 76". Area = 
16.6 sq. ins. 

3, In Ex. 2, Art. 156, find the average school attendance for the 

six years 1892-97. Ans. 119,100. 



46 PRACTICAL PLANE GEOMETRY chap. 

44. Miscellaneous Examples. 

Note. The figures are to be copied twice size. 

*1. Divide the given triangle ABC into four equal parts by lines 

drawn parallel to AB. (1896) 

*2. Draw a triangle similar to the given triangle ABC, but with a 

perimeter equal to the harmonic mean between the lines LAI 

and PQ. (1896) 

*3. Divide the given line A into two parts, such that the sum of the 

squares on them shall be equal to the square on the given line 

B. State what limitations are necessary as to the length of B 

in order that the problem may be possible. (1S88) 

Hint. The solution of a problem is often suggested by an 

algebraical investigation. 

Thus let x be one part, and therefore A - x the other. Then 
by the conditions x 2 + (A - x) 2 = B 2 . Solving which, we find 

X = iJyA Ji^Bf-A*}. 

Now V2.5 is the diagonal of a square on the side B. If 
this be made the hypothenuse of a right-angled triangle, and 

A one of the sides, the other side is / (V ' 2B) 2 - A 2 . The 

rest of the construction is simple. 

*4. Between the lines ab, cd produced if necessary, place a line 
parallel to ef in such a position as to form a four -sided figure 
with an area of 2 square inches. (r89i) 

Hint. Produce ab, cd to meet in o. Find the area of the 
triangle ofe. Call this m square inches. Then find the triangle 
oFE similar to ofe, the areas of the two having the ratio 2 + m : m. 

*5. Draw a line ab parallel to the base AC of the given triangle 

ABC, cutting off segments Aa on AB, Cb on CB, such that 

their difference Aa Cb shall be equal to one inch. ( 1S95) 

Hint. Aa may be found as the fourth term in the following 

proportion 

AB-BC :AB:: Aa - Cb : Aa. . 

*6. Construct an equilateral triangle which has one angular point at 
b on the line ob, and the remaining angular points on oa and oc 
respectively. (H. 1886) 

7. Draw a triangle ABC with the following dimensions. AB = ^", 

BC=2^", AC = 2". Inscribe in the triangle a rhombus, one 
side lying in AC, and the adjoining sides inclined to AC at 

45- (1894) 

8. Given two lengths P= 2. 25", Q=2."jo". Find a line whose 

length x is such that P 2 = (x Q) x. Write down the value of x. 

(1894) 



ii SIMILAR RECTILINEAL FIGURES AREAS 47 




4 8 PRACTICAL PLANE GEOMETRY chap. 

Hint. P is a mean proportional between (x - Q) and x. 
Therefore in Fig. 27, page 33, make OB = P, OC=Q. Draw 
CA perpendicular to CB. Then AB = x. 
9. In Ex. 8 find x if P 2 = (Q-x)x. 

*10. Reduce the given hatched figure to a triangle. (1883) 

Hint. Reduce the whole figure and the hole to triangles. 
Then determine a triangle equal to the difference of the two. 

*11. Through the given point p draw two lines cutting the two 
given lines in such a way that the included area is i\ square 
inches. (!887) 

Hint. Through /draw any two lines intersecting the given 
lines in a, A ; b, B. Reduce AabB to a square, and find its 
area in square inches by Prob. 38. The length of AB required 
for an area of 3^ square inches may now be found as a fourth 
proportional to area AabB, 3j, AB. 

*12. In what sense can areas be represented by straight lines ? If 
the area of the given triangle ABC is represented by a line 1" 
long, draw a line representing the area DEFG. (1S85) 

*13. Reduce the given figure to a rectangle on the base ab. (1886) 

Hint. Apply a construction similar to that of Prob. 30. Thus 

drawy?, Im, mn respectively parallel to ge, gd, gc, to meet the 

lines ed, dc, ca. Join gn, and through its middle point draw a 

line parallel to ab. We thus obtain the required rectangle. 

*14. Draw a straight line from A and terminating on FG, such 
that the sum of the areas included between it and the given 
zigzag line on the one side is equal to the sum of those on the 
other. (1885) 

15. Draw a quadrilateral figure having an area of 3 square inches, 

and made up of two isosceles triangles having a common base 
which is a mean proportional between the sides of the triangles. 
The vertical angle of the smaller triangle is 90 . (1881) 

16. A line 1-5" long represents a square of 2" side. Determine a 

length of line which would represent on the same scale a 
hexagon of if" side. ( x 883) 

17. Draw a right-angled triangle with hypothenuse 2" and one side 

if". Prove by construction that the square on the hypothenuse 

is equal to the sum of the squares on the sides. ^893) 

Hint. One method is to apply the construction of Prob. ^8. 

18. Draw an equilateral triangle of 2" side, and circumscribe this by 

a right-angled isosceles triangle, so that one of the equal sides 
of the latter is bisected. 



ir SIMILAR RECTILINEAL FIGURES AREAS 49 






Cb/ii/ -tfie figures double size 



p 




E 



CHAPTER III 

TRIANGLES, CIRCLES AND LINES IN CONTACT 

45. Introduction. The problems in this chapter relate 
mainly to the contact of lines and circles, and the con- 
struction of triangles from adequate data. 

Euclid permits a line to be drawn between two definite 
points, but does not allow a common tangent to be drawn 
to two given circles without having previously determined 
the points of contact by a special construction. 

This construction may be necessary in a strict system of 
deductive reasoning like Euclidean geometry, but does not 
add to the accuracy in a scale drawing, as the student may 
easily test for himself. 

So in "practical" geometry a tangent may be drawn to 
a circle from an external point, or a common tangent to 
two circles, by simply adjusting the straight-edge and draw- 
ing the line without any previous construction. Then if 
the point of contact be required, it is necessary to draw the 
perpendicular radius. 

If a tangent is required at a point on the circumference, 
it must be drawn perpendicular to the radius to the point. 

With care, a circle may be drawn with a given centre to 
touch a given line, without first finding the point of contact. 
But in this case it is generally preferable to draw the 
perpendicular from the point to the line before drawing the 
circle. 



ch. in TRIANGLES, CIRCLES AND LINES IN CONTACT 51 

Before working the problems of this chapter, the student 
should illustrate the truth of the following theorems, by 
making accurate drawings to scale, measuring the results 
and making calculations where necessary. For several 
reasons this is a most valuable form of exercise ; by com- 
paring his results with the true ones he may observe how 
small need the errors be which he introduces into his 
graphical work. It will not be necessary to add that he 
must see that his pencil is in proper condition. 

Theorem 1. Angles in the same segment oj a circle are 
equal to each other (Euc. III. 21). 

Theorem 2. The two tangents drawn from a point to a 
circle are equal to each other (Euc. III. 17). 

Theorem 3. If through any point in the circumference of 
a circle a chord and tangent be drawn, the angles between 
them arc equal to the angles in the alternate segments of the 
circle (Euc. III. 32). 

Theorem 4. If any point P be taken inside or outside a 
circle and two lines be drawn through P, one cutting the circle 
at A and B, and the other at C and T>, the product of PA 
and PB is equal to the product of PC and PD (Euc. III. 35). 

Also if Pf>e outside the circle and a line PT be drawn to 
touch the circle at T, the squa?'e of PT is equal to the pro- 
duct of PC and PD, or of PA and PB (Euc. III. 36). 

Theorem 5. If a line which bisects the vertical angle A 
of any triangle ABC, cut the base BC in D, the ratio of BD 
to DC is the same as the ratio of BA to AC (Euc. VI. 3). 

Also if a line bisecting the exterior angle at A cut the 
base BC produced in D\ the ratio of D'C to D'B is the 
same as that of AC to AB (Euc. VI. 3). 

D and D' are said to divide BC internally and externally 
in the ratio of the sides of the triangle. (Or AB, AD, AC, 
AD' form a harmonic pencil. See Prob. 29.) 

Theorem 6. If A and B are two fixed points, and a 
poitit P move so that the ratio PA : PB is constant, then the 
locus of P is a circle. 



5 2 PRACTICAL TLANE GEOMETRY chap. 

46. Problem. On a given line AB, to describe a 
regular polygon ; say a heptagon. 

First Method. Produce AB, and with centre B, radius 
BA, describe the semicircle AC. 

By trial, divide the semicircle into as many equal parts 
as the polygon has sides, in this case seven. Join B to 
the second point of division from C. 

Then B2 is a second side of the required heptagon. 

Draw a circle through A, B, 2. This is the circum- 
scribing circle of the required polygon, and the length AB 
should step exactly seven times round the circumference. 

Second Method. Produce AB, and set off the angle CB2 
equal to 360-=- 7, that is to 51. 4 . 

Make B2 equal to BA. Then proceed as before. 

Note. To inscribe a regular polygon of n sides in a circle, divide the 
circumference by trial with the dividers into n equal parts ; or, find one 
side by setting off from the centre an angle equal to 360-=- n. 

47. Problem. On a given line AB, to construct a 
segment of a circle which shall contain an angle equal to 
a given angle a. 

Draw CD bisecting AB at right angles, and make the 
angle DCE equal to a. Draw AO parallel to CE. 

Describe a circle with O as centre, radius OA ; then 
that segment of the circle, on the side of AB on which D 
lies, is the one required. 

If the given angle is a right angle, the point O coincides 
with C. If the angle is obtuse, O is on DC produced. 

48. Problem. From a given circle to cut off a segment 
which shall contain an angle equal to a given angle. 

Take a point A on the circumference and draw the 
tangent AD. 

Make the angle DAB equal to the given angle. 
Then the segment ACB is the one required. 

49. Problem. In a given circle to inscribe a triangle 
the sides of which shall be in a given ratio. 

Draw any triangle having its sides in the given ratio. 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 53 




At any point A in the circumference of the given circle 
draw the tangent DE. Make the angles DAB, EAC 
respectively equal to two of the angles of the triangle, 
join BC. Then the triangle ABC is the one required. 

50. Problem. To construct a triangle, having given 
the base, vertical angle, and altitude. 

Let BC be the given base ; describe on it the segment 
of a circle containing an angle equal to the given vertical 
angle. Draw DE parallel to BC, the distance between 
these lines being equal to the given altitude. If A, A' are 
the points in which DE intersects the segment, either of 
the triangles ABC, A' BC will satisfy the given conditions. 



54 PRACTICAL PLANE GEOMETRY chap. 

51. Problem. The perimeter of a triangle is 5" ; the 
vertical angle is 70 ; and one of the sides is half the hase. 
Construct the triangle. 

Take any line AB ; on AB draw a segment of a circle 
containing an angle of 70 (Prob. 47). 

With centre A, radius half AB, describe an arc cutting 
the circle in C. Join CA, CB. 

Then ABC is a triangle similar to the one required. 

Produce AB both ways; make AE equal to AC, and 
BD equal to BC. 

Divide a line 5" long into segments having the ratio 
EA : AB : BD (Prob. 8). Then these segments are 
equal to the required sides of the triangle. 

52. Problem. To construct a triangle, having given 
the base, one of the hase angles, and the perimeter. 

Let BC be the given base. Draw BE such that the 
angle CBE is equal to the given base angle. Along BE 
set off BD equal to the sum of the remaining sides. 

Join DC ; draw EA bisecting DC at right angles, and 
join AC. Then the triangle ABC is the one required. 

53. Problem. To construct a triangle, having given 
the base 2", perimeter 5", and area 0-85 square inch. 

In the method here given we first determine the side of a square 
0.85 square inch in area, and then obtain a rectangle equal in area to 
this square, one side of the rectangle being equal to the given base, 2". 
The other side of this rectangle will be equal to half the altitude of the 
required triangle. 

Make OS=i" and <9iV= 0.85". On SN describe a 
semicircle, and draw OX perpendicular to SJV; then OX is 
the length of the side of the square in question. 

Now make OR- given base =2". Bisect RX at right 
angles by the line YO\ and with O' as centre and radius 
O'X describe the arc XT; then the area of the rectangle 
RO.OT= (OX) 2 = 0.85 square inch, and therefore OT= 
half altitude of required triangle. The problem may now 
be completed as in Prob. 55. 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 55 




R 



51 



D,a 




x 



YJ" 



S 0' 



53 



D 



T N 



Examples. 1. Describe a regular pentagon on a line if" long. 

2. In a circle 2|" diameter inscribe a triangle two angles of which 

are 45 and 65 . 

3. Construct a triangle having the base if, perimeter 4|", and 

area .95 of a square inch. 



56 PRACTICAL PLANE GEOMETRY chap. 

54. Problem. To construct a triangle, having given 
the vertical angle, altitude, and perimeter. 

Draw two lines AD, AE containing an angle equal to 
the given vertical angle. With centre A, and radius equal 
to the given altitude, describe the arc MN, 

Along AD and AE set off AF and AG each equal to 
half the given perimeter. 

Draw FO and GO perpendicular to FA and GA, and 
with as centre, describe the arc GF. If now a com- 
mon tangent be drawn to this arc and the arc MN, so 
as to meet AD and AE in B and C respectively, then 
ABC is the required triangle. 

55. Problem. To construct a triangle, having given 
the base, altitude, and the perimeter. 

Let F'Fbe the given base; bisect F'F in C, and make 
CA = CA' = half the sum of remaining sides, i.e. = ^ (peri- 
meter - base). Draw CE perpendicular to AA', and with 
T^as centre and CA as radius, describe an arc intersecting 
CE in B. Make CQ = given altitude, and CG = CB. 
Join GQ, and draw A' K parallel to GQ. On A A' describe 
a semicircle, and draw KP' parallel to A' A. Draw F'F 
parallel to BC, and QF parallel to KP' , then PFF' is the 
required triangle. 

Note. The point P is on an ellipse, major axis A A', foci F, F'. 
The above construction is based on the properties of an ellipse. 

56. Problem. To construct a triangle, having given 
H, K, L, the lengths of the three medians. 

A median is a line drawn from any vertex to the middle 
point of the opposite side. 

Make AD = Ff, and produce it to G such that DG = 
DO = IAD. 

^'ith O and G as centres, and f K and L as radii 
respectively, describe two arcs intersecting at B. 

Join BD and produce it to C, making DC ' = DB, then 
ABC will be the required triangle. 

Note. For explanation, compare the construction with the triangle 
abc in which the medians have been drawn. 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 57 








56 



-JG' 



58 PRACTICAL PLANE GEOMETRY chap. 

57. Problem. To construct a triangle, having given 
(a) the hase, vertical angle, and the ratio of the sides, 
say 3:5; (h) the base, altitude, and the ratio of the 
sides, 3 : 5. 

(a) Let AB be the given base. Select any convenient 
unit, and construct the triangle AP'B, making BP' = 3 
units, and AP' = 5 units. Bisect the angle AP'B by the 
line P'D, and draw P' D' perpendicular to P ' D. 

On DD' as diameter describe a circle ; this circle must 
contain the vertex of the required triangle. On AB 
describe the segment ABE of a circle, containing an angle 
equal to the given vertical angle (Prob. 47). Then the point 
P in which this segment intersects the circle on DD' is 
the required vertex, and APB the required triangle. 

(b) Proceed as in (a) so far as obtaining the circle on 
DD'. Draw AF perpendicular to AB, making AF= the 
given altitude ; draw FG parallel to AB, intersecting, in 
P, the circle on DD', then APB is the required triangle. 

The construction for this problem is based on Theorems 5 and 6, 
Art. 45. 

58. Problem. To construct a triangle equal in area 
to a given triangle ABC, and having two of its sides 
respectively equal to two given lines D and E. 

Draw AF perpendicular to BC, bisect AF in G, and 
complete the rectangle BHKC. Draw LM equal to one 
of the given lines, say D, and mark off LN equal to BC. 
Draw MO perpendicular to ML, making MO equal to 
FG ; join LO, and draw NP parallel to MO, then the 
rectangles QRML and BHKC are equal in area (Prob. 
34). Produce NP to ,.9, making PS equal to PN, and 
draw ST parallel to LM. With L as centre, and the 
other given line E as radius, describe an arc intersecting 
TS in V, then the triangle VLM satisfies the required 
conditions. 

Note. The arc described with L as centre would intersect ST 
produced in V, hence there are two solutions. 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 59 



E 




A\ 2K IB 



D 




N M 



Examples. 1. Construct a triangle having the vertical angle 
55j altitude 2", and perimeter 5 inches. 

2. Construct a triangle having the base 2'', altitude \\" , and 

perimeter 5 inches. 

3. The three medians of a triangle are i|", 2j", and 3" ; draw the 

triangle. 

4. Construct a triangle having its base 3", vertical angle 120, and 

the ratio of sides 3 : 5. 

5. Construct a triangle having its base 2", altitude 2j", and ratio 

of sides 4 : 7. 

6. Draw a triangle with sides of 2", 2\", 2|", and construct another 

triangle having the same area, two of its sides being 2" and 3". 



60 PRACTICAL PLANE GEOMETRY chap. 

59. Problem. To describe a circle to pass through two 
given points P and Q, and touch a given line AB. 

Join P and Q, and produce PQ to cut AB in P. 

Along AB set off R T (either to the right or left) equal to the mean 
proportional between PQ and PP. 

Draw TD at right angles to AB, and bisect PQ at right angles by 
NO, meeting TD in 0. 

Then is the centre of one circle satisfying the required conditions. 

60. Problem. To describe a circle which shall have 
its centre on a given line CD, pass through a given point 
P, and touch a given line AB. 

Let fall a perpendicular /Wfrom P on to CD ; and produce PN 
to Q, making NQ equal PN. Then, by Prob. 59, determine a circle 
to pass through P and Q and touch AB. 

61. Problem. To describe a circle to pass through a 
given point P, and touch two given lines BA, AC. 

Draw AD bisecting the angle CAB. 

With any point O 1 in AD as centre, describe the circle touching 
AC and AB. 

Draw AP, and produce it to cut this circle in P' (and P 1 not shown). 
Join P'O, and draw PO parallel to P'O. 

Then O is the centre of one circle satisfying the conditions. 

62. Problem. Between two given lines AB, AC, to 
inscribe a succession of circles in contact. 

Draw AD to bisect the angle BAC. Take any point O on AD as 
centre, and describe a circle touching AB and AC, and cutting AD in 
E, F. 

Draw OG perpendicular to AB, and join EG, FG. Draw EH, 
HP respectively parallel to FG, GO ; and FK, KQ to EG, GO. 

With centre P and Q, describe circles through E and F. 

The three circles will touch the two lines and each other as required. 
The circles may be extended by repeating the construction. 

63. Problem. Two straight lines AB and CD are 
given, their point of intersection being inaccessible. It 
is required to describe a circle which shall touch these 
lines and pass through a given point P. 

By Prob. 22, draw FF to bisect the angle between AB and CD. 
With any centre O in EF, draw the circle which touches AB. 

By Prob. 24, draw PQ to converge to the same point as AB, CD, 
cutting the circle in P and P. Draw PO' parallel to PO. Then the 
circle through P, with centre O' , is one solution. 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 61 



Q D, 




62 K 




62 PRACTICAL PLANE GEOMETRY chap. 

64. Problem. To describe a circle which shall pass 
through two given points P and Q, and touch a given 
circle, centre G. 

Join PQ, and draw NK bisecting PQ at right angles. 

Select any point C on NK such that the circle, with C as centre 
and CQ as radius, will intersect the given circle in E and F say. 

Join FE, and produce it to meet PQ produced in R. Through R 
draw the tangent RT (or RT'), and produce GT (or T'G) to meet 
NK in (or 0') ; then O (or O') is the centre of a circle .satisfying 
the given conditions. 

For RT 2 =RE.RF=RQ. RP {Euc. III. 36). 

Hence the circle through P, Q, T touches RT at T (Euc. III. 37), 
and therefore also the given circle at T. Consequently the centre of 
the required circle lies on GT produced (converse of Euc. III. 12) ; 
but it must lie on NK {Exxc. III. 1) ; hence it must be at O (or G J ). 

65. Problem. To describe a circle which shall have 
its centre on a given line CD, pass through a given point 
P, and touch a given circle, centre G-. 

Let fall a perpendicular PN from P on to CD, and produce PN to 
Q, making NQ equal PN 

Then, by Prob. 64, determine a circle to pass through P and Q, 
and touch the given circle, centre G. 

66. Problem. To describe a circle to touch two given 
circles, centres A, B, and pass through a given point P. 

Draw a common tangent Tt to meet AB produced in S ; join PS. 
Through G, F, P describe a circle cutting PS in Q. 

By Prob. 64 describe a circle to pass through P, Q, and to touch the 
circle with centre B ; D being the point of contact. (There are two 
such circles.) This circle will also touch the circle with centre A. 

Also, by drawing the internal tangent to the given circles, meeting 
AB in S', two others may be found by a similar construction. 

67. Problem. To describe a circle which shall touch 
a given circle, centre Gr, pass through a given point P, and 
touch a given straight line AB. 

Draw GE perpendicular to AB, meeting the given circle in C and 
D ; join CP. 

Describe a circle to pass through P, D, E, cutting PC in F. 

By Prob. 59 describe a circle passing through P, F, and touching 
AB ; this will be the required circle. 

Note. There are two circles passing through P, F and touching AB. 

Also, by reading D for C, and C for D, in the above instructions, 
two other circles will be found to be possible. 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 63 




64 PRACTICAL PLANE GEOMETRY chap. 



68. Problem. To describe a circle which shall touch 
three given circles, centres A, B, C. 

Let r v r 2 , r 3 denote the lengths of the radii of the circles, centres 
A, B, C respectively, the first being the least. 

With centre B and radius r., - i\, describe a circle, and with centre 
C and radius r 3 - r v describe another circle. 

By means of Prob. 66 determine 0, the centre of a circle which 
passes through A and touches the two circles just described. Join 
OA, OB, OC, meeting the circles at the points indicated. This 
construction ensures that AD=EG-FH; from which it follows that 
the circle described with O as centre and OD as radius will touch 
three given circles externally. 

In general there will be eight circles, satisfying the given condi- 
tions ; three being such that for each one there is internal contact with 
two of the given circles, and external contact with the third ; three 
others which have external contact with two of the given circles and 
internal contact with the third ; one circle which has internal contact 
with the given circles ; finally, the one in the figure. 

69. Problem. To describe a circle which shall touch 
two given lines AB, AD, and a given circle, centre C. 

Draw EF parallel to AD, and GH parallel to AB, the distance 
between the parallel lines in each case being equal to the radius of 
the given circle. 

Now draw a circle which shall pass through C and touch EF and 
GH, Prob. 6 1 (there are two such circles) ; let be its centre, Join 
OC, cutting the given circle in T. 

Then the circle with centre O and radius OT is one solution. 

Note. If EF and GH be drawn on the other side of AD and 
AB respectively, a second solution will be obtained, the circles having 
internal contact. 

70. Problem. To describe a circle which shall touch 
a given line AB and two given circles, centres C, D. 

Draw EF parallel to AB, at a distance equal to the radius of one 
of the other circles, say the one with centre C. With centre D describe 
a circle whose radius is equal to the difference of the radii of the given 
circles. 

By Prob. 67 describe a circle to touch the latter circle, pass through 
C and touch EF; let O be its centre. Join OC, cutting the circle, 
centre C, at G ; then the circle with as centre and OG as radius 
will satisfy the required conditions. 

Note. There will, in general, be eight solutions. In four cases 
EF will lie on one side of AB, and in the other four cases on the 
opposite side. 



ni TRIANGLES, CIRCLES AND LINES IN CONTACT 6 








F 



66 PRACTICAL PLANE GEOMETRY chap. 

71. Problem. To describe a circle which shall touch a 
given circle, centre G, and a given line AB at a given 
point T. 

1st. Externally. Draw TF z.\\A GD perpendicular to AB. 
Join DT, intersecting the given circle at E. 
Produce GE to : then O is the centre of the required circle. 
2nd. So as to include the given circle. Proceed as above, reading 
D', ', and 0' for D, E, and 0. 

72. Problem. To describe a circle which shall touch 
a given line AB, and a given circle, centre G, at a given 
point E. 

Join GE, and produce it both ways. 

Draw the tangent ER, and set off R T and RT' , each equal to RE. 

Draw TO and T'O', each perpendicular to AB ; then and 0' 
are the centres of the two required circles, having respectively external 
and internal contact with the given circle. 

73. Problem. To describe a circle which shall have 
its centre on a given line CD, and shall touch a given line 
AB and circle, centre G. 

By a method illustrating the use of a locus. 

Draw any line Jl/JV at right angles to AB, and any line GH, 
cutting the given circle in K. 

Along MN set off any lengths Ml, Mz, M$, etc., and along KH 
set off A'l, A~2, and A'3 respectively equal to these. 

With G as centre describe arcs through the points I, 2, 3, etc., on 
A'AA to meet lines drawn parallel to AB through the corresponding 
points 1, 2, 3, etc., on MN. 

These arcs and corresponding lines intersect at points 1, 2, 3, etc., 
through which points draw a fair curve. 

This curve is the locus of the centre of a circle which moves so as 
always to touch the given line and circle, and the point in which 
the curve intersects CD will be such that the circle described with 
as centre, to touch AB, will also touch the circle, centre G. 

Examples. Take two points A and B 2" apart. Draw a line 
CD distant 1" from A and i|" from B. On CD take P, i|" 
from A. Describe a circle which shall pass through 

1. A and B and touch CD. 

2. A, touch CD, and have its centre on AB. 

3. A and B, and touch a circle, centre P, radius \" . 

4. A, have its centre on AB, and touch circle centre P, radius i". 

5. B, touch CD, and a circle, centre A, radius f". 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 67 




73 



68 PRACTICAL PLANE GEOMETRY chap. 

74. Miscellaneous Examples. 

*1. In the given circle place a chord 1.8" long and parallel to the 
line joining the given points /, q. Draw a second chord of the 
same length which, if produced, will pass through/. (1884) 

Hint. If two circles are concentric, all chords of the outer 
circle which touch the inner are equal. 

2. Draw two lines including 50 . Describe a circle 2^" diameter 

cutting these lines in chords of I-|" and 2|". (1893) 

3. Two points a, b are 3|- miles apart. Determine the position 

of a point/, such that pa is l miles, and the angle apb 73. 

Scale l"=lj miles. (1886) 

*4. From a draw a line cutting the two given circles in equal chords. 

N.B. An auxiliary curve may be employed. ('893) 

Hint. Draw any line through a cutting the circles in chords 
ab,cd; along cd mark off ex = ab. Repeat this construction for 
other lines through a, and draw a fair curve through the points x. 
Let the locus of x cut the circle in X \ then aX is one solution. 

"*5. Through A draw two lines containing an angle of 30 , and 
intercepting a length of 2" on the given line BC. (1889) 

*6. The figure represents a window of the decorated English 
style. The construction is sufficiently shown and dimensions 
are given. Draw the window to scale of 2" to 1' o". (1879) 
7. Describe a circle enclosing and touching three other circles of 
0.6", 0.8", and 1" radius respectively, each of which touches 
the other two. (1882) 

*8. Draw the riband pattern to the given dimensions. (1880) 

9. Three posts, B, C, D, are in a straight line at intervals of 100 

yards. An observer at A finds that the angle BAC is 20 , and 

CAD 30 . Obtain the position of A (Scale TT Vff)- (1882) 

10. Construct a regular pentagon having a diagonal of 3". (1881) 

11. Draw a triangle from the following data : 

(1) Base 2j", perimeter 7 J", one base angle 55 . (1880) 

(2) Perimeter 6", vertical angle 6o, altitude 1.25". (1890) 

(3) Altitude 2.5", perimeter 8J", one base angle 50. (1877) 
*12. A boy starting from a runs in the direction ab. A man starts 

at the same moment from c. Supposing the man to run also in 
a straight line, but 1^ times as fast as the boy, what direction 
must the former take to catch the latter? (1880) 

*13. Describe a circle of 2^" radius touching the two given circles. 
The given circles are to be within the required circle. (1877) 

*14. Describe three circles touching the two given lines AB, CD, 
and each other successively 

(1) Such that the middle circle passes through P. (1878) 

(2) Such that the largest has a radius of 1 inch. (1881) 



in TRIANGLES, CIRCLES AND LINES IN CONTACT 69 




CHAPTER IV 



CONIC SECTIONS 



75. Definitions. Let one of two intersecting straight 
lines of indefinite length revolve about the other as a fixed 
axis, the inclination to each other and the point of intersec- 
tion of the lines remaining fixed, then the two equal and 
opposite acute angles generate two equal and opposite conical 
solid angles of revolution, or a double cone of unlimited 
extent. The revolving line generates the surface of the 
cone, and in any position is called a generator of the 
surface, or simply a generator. The fixed line is called 
the axis, and the point of intersection the vertex, of the 
cone. 

Any plane section of this cone is called a conic section. 

The object of the present chapter is to consider the 
nature of these conic sections together with some of their 
more useful geometrical properties, and to give convenient 
methods of constructing the curves. 

Classification of conic sections. If the section plane 
be at right angles to the axis and do not contain the vertex, 
the conic section is called a circle. 

If the section plane be not at right angles to the axis, 
nor parallel to a tangent plane, and be such that (supposed 
indefinitely extended) it cuts only one of the halves of the 
double cone, the section is called an ellipse. 

If the section plane be parallel to one generator, and 



chap, iv CONIC SECTIONS 71 

one only, that is, be parallel to a plane which touches the 
cone, the conic section is called a parabola. 

If the section plane cut both halves of the double cone, 
and do not contain the vertex, the section is called an 
hyperbola. 

If the section plane pass through the vertex, the conic 
section will be either a point, a line, or two plane angles 
opposite to each other. These may be considered as 
limiting cases of the ellipse, parabola, and hyperbola 
respectively. 

From the manner in which the above conic sections 
are formed, it will be obvious that both the circle and 
ellipse are completely bounded figures, while the figure of 
the parabola extends indefinitely in one direction, and that 
of the hyberbola consists of two portions, which extend 
indefinitely and in opposite directions. These remarks 
refer merely to the general appearance of the sections ; we 
have now to consider other relations which distinguish these 
figures. 

Double meaning of the ivords "cone" and " conic section." 
A circle, as defined by Euclid, is the plane figure 
enclosed within the circumference, not the circumference 
itself, though the word " circle " is frequently used as 
synonymous with the latter. The terms "ellipse," "para- 
bola," and "hyperbola" are also used in two senses, 
denoting in one case a plane figure, and in the other the 
bounding curve or outline of the figure. In like manner 
the word " cone " is commonly employed with a double 
meaning, indicating either the solid figure or its bounding 
surface. So the sections of a cone may mean either the 
plane figures or their curved outlines. 

In this book we shall have occasion to use the terms 
in both senses ; the reader must infer the meaning from the 
context in any particular case. 

In the next article the conic sections are examined from 
another standpoint, being regarded as curves traced by a 
moving point ; and the nature of the motion is defined. 



7 2 PRACTICAL PLANE GEOMETRY chap. 



76. Properties of conic sections. Let V be the vertex, 
and VO the axis of the cone. DP represents a section 
plane which determines the conic section PAQ. Let a 
sphere, centre /, be inscribed in the cone so as to touch 
the plane at F. This sphere touches the cone in a circle, 
centre O, the plane of which {HK) is perpendicular to the 
axis ; let this plane intersect the section plane in the line 

DN. 

Select any point P on the curve PAQ; take PM per- 
pendicular to the plane HK, meeting it in M; draw PN 
perpendicular to DN, and join PV, PP, ML, MN. In 
this way two right-angled triangles PML, PMN are formed, 
since PM, being perpendicular to the plane HK, is there- 
fore perpendicular to the lines ML, MN in the plane. 

Now the angle PLM is equal to the complement of the 
semi-vertical angle of the cone, and is therefore constant 
for any position of P, hence the ratio PL : PM is constant. 

Again, the angle MNP, or a, is equal to angle between 
the planes HK, DP and is therefore constant, hence 
the ratio PN:PM\s constant, 

therefore the ratio PL : PN is constant. 

But PL and PP are tangents from P to the same 
sphere, therefore PL = PP, hence 

the ratio PP: PN is constant. 

That is to say, wherever P may be on the curve PAQ, its 
distance from F is always in a constant ratio to its distance 
from DN, and this is true for the parabola, ellipse, and 
hyperbola. 

The point F is called a focus, the inscribed sphere a 
focal sphere, the line DN is a directrix, and the ratio 
PF: /Wis called the eccentricity of the conic section. 
Thus we obtain the following definition : 
Definition. A conic section is the carve described by a 
point which moves in a plane in such a manner that its dis- 
tance from a fixed point in the plane (a focus) is in a constant 
ratio to its distance from a fixed line (a directrix) in the plane. 



IV 



CONIC SECTIONS 



73 




Referring again to the figure, the triangle PLM may- 
be supposed to rotate about PM until its plane coincides 
with that of the triangle PMN, when LAIN becomes one 
straight line. The triangles will then appear as shown 
detached at P Q L Q M N. The left-hand diagram of tiie 
figure is an elevation on a plane containing the axis of the 
cone and perpendicular to the section plane. In these 
diagrams corresponding points and angles are denoted by 
corresponding letters. 

From the manner in which the curves were defined in 
Art. 75, it is obvious that for an ellipse the section plane 
must be so chosen that a is less than ft, hence P L is less 
than jPqNq, or PF is less than PN. For a parabola a is 
equal to ft, hence PF is equal to PN. And for a hyperbola 
a is greater than ft, and consequently PF greater than PN. 

Hence it follows that a conic section is an ellipse, a 
parabola, or a hyperbola, according as its eccentricity is 
less than, equal to, or greater than unity. 



74 PRACTICAL PLANE GEOMETRY chap. 

77. The ellipse. Let the cone be cut by the plane 
DD' so that the section is an ellipse. A focal sphere is 
shown touching the section plane in a focus F, and deter- 
mining the directrix FN, as explained in Art. 76. A 
second inscribed sphere is shown on the other side of the 
cutting plane touching the latter in F'. D ' N' is the line 
of intersection of the cutting plane with the plane contain- 
ing the circle of contact H'L'K'. 

Take any generator meeting the ellipse in P, and the 
circles of contact in Z, F. Let" JZ/W be a line parallel 
to the axis -of the cone. Draw NPN' perpendicular to 
ZW or FN'. Join ML, MN, M 'L ',. MN', FF, FF '. 

Suppose the triangles LMP, L'M'F are turned about 
MM' into one plane as shown detached to a reduced scale, 
then it will be readily seen that 

FF : FN= FF' : FN' = PL : FN, or PL' : FN. 

Therefore F is similarly related to F' and D ' N' as to F 
and FN; thus the ellipse has two foci, F, F', and two 
directrices, FN, F'N. 

Let the line through FF meet the curve in A, A'. 
Bisect FF' in C and draw BB' perpendicular to A A'. 

Since the curve may be denned with reference to either 
the focus Zand directrix FN, or the focus F 1 and directrix 
D ' N', it follows that BB' is an axis of symmetry ; so also 
is A A'. Therefore the ellipse has a centre which is C. 

Any line through C and terminated by the curve is 
called a diameter ; and all diameters are bisected at C. 

A A' is called the ma/or axis, and BB' the minor axis ; 
they are respectively the longest and shortest diameters, 
and are sometimes referred to as the prituipal axes. 

It appears from the figure that 

PF+ FF' = PL + PL' = LCLC = AF+ AF' = A A'. 

Theorem 1. Ln any ellipse the sum of the focal distances 
is constant and eqital to the major axis, or to the length of 
the portion of a generating line of the cone intercepted by the 
circles of contact of the focal spheres. 



IV 



CONIC SECTIONS 



75 



L M 




Theorem 2. // may be shown that CF, CA, CD are i>i 
geometrical progression ; also that the eccentricity FA : AD 
= CF: CA = CA: CD. 



76 PRACTICAL PLANE GEOMETRY chap. 

78. Problem. Having given the lengths of the major 
and minor axes of an ellipse, to determine the foci. 

Let AA' be the major axis. 

Bisect AA' in C ; draw CB, CB' at right angles to AA', 
making each equal to half the minor axis. 

With B or B' as centre and CA or CA' as radius, de- 
scribe arcs cutting A A' in i^and F ' . 

Then Fand F' are the required foci. 

79. Problem. Having given the foci and the major 
axis of an ellipse, to determine the minor axis. 

Let F, F' be the two foci and AA' the major axis. 
Bisect AA' in C. With Fand F' as centres and radius 
CA describe arcs to cut each other in B and B'. 
Then BB' is the minor axis. 

80. Problem. To construct an ellipse, having given 
the major and minor axes AA', BB'. (First method.) 

Determine i^and F', the foci, as in Prob. 78. 

In CF' take any convenient points 1, 2, 3 . . . 

With A 1 as radius describe arcs with centres F and F' 
respectively; similarly with A'l as radius describe arcs with 
centres F' and F. These arcs intersect in the four points 
marked 1,. 

Repeat this construction for the points 2, 3, thus obtain- 
ing the points marked 2 V 3 r 

Then the points i v 2 V 3 X are on the required ellipse. 

A fair curve should now be drawn through the points 

'i> 2 v 3]/ 

81. Mechanical method of describing an ellipse. 

Insert pins at F, F', B. Take a piece of fine string, 
pass it round the pins so as to form a triangle F'BF, and 
tie the ends tightly together. Replace the pin B by a 
sharply-pointed pencil, with which trace out the curve, 
allowing the pencil to be guided by the string, but not 
pressing it against the string too tightly. 



IV 



CONIC SECTIONS 



77 







B 


> 




A'l F / 2 


3, 
3 




A ** 

7 

c 


A 


X 

V^j 


J 




3, 


Zi 


SI 



78, 79, 80 




F A 



Examples. 1. The lengths of major and minor axes of an ellipse 
are 4" and 3" respectively; determine and measure the dis- 
tance between the foci. A us. 2.64". 

2. The major axis of an ellipse is 3.6" long, and the foci are 2.3" 

apart ; find the length of the minor axis. Ans. 2.78". 

3. The minor axis of an ellipse is 2.8" long, and the foci are 2" 

apart ; find the length of the major axis. Ans. 3.44". 

4. Construct the ellipse of Ex. 1 by the method of Prob. 80. 

5. Trace the ellipse of Ex. 3 by the method of Art. Si. 



j8 PRACTICAL PLANE GEOMETRY chap. 

82. Projective properties of the ellipse. It may be 
proved that any orthogonal or radial projection of a conic 
section is a conic section. It will be sufficient here to state 
this theorem in the less general form : 

The orthogonal projection of an ellipse is an ellipse. 

In this theorem the ellipse is to be considered as in- 
cluding the circle and the straight line as extreme cases. 

By considering an ellipse as the projection of a circle, 
and a circle as the projection of an ellipse, we shall be 
able to prove in a simple manner some useful properties of 
the curve. In this connection two additional theorems of 
projection will be required. 

A system of parallel straight lines project orthogonally 
into a system of parallel straight lines, the lengths of the pro- 
jections bearing the same ratios to each other as the lengths of 
the corresponding lines bear to each other. 

If a line be tangential to a curve at any point, then any 
projection of the line will be tangential to the projection of the 
curve at the projectio?i of the point. 

Let AA', B Q B ' be two diameters of a circle perpen- 
dicular to each other. 

Suppose the circle to be turned about AA' as axis until 
the projection of B B Q ' on the plane of the paper is BB' ; 
then the projection of the circle will be the ellipse shown, 
of which AA ', BB' are the principal axes. 

Next conceive the ellipse, the plane of which is that of 
the paper, to be turned about the minor axis until the major 
axis projects into the line A X A^ of length equal to BB' ; 
then the projection of the ellipse will be the circle with 
BB' as diameter. 

Definitions. The circle described on the major axis 
of an ellipse as diameter is called the major auxiliary 
circle, or simply the auxiliary circle or the major circle. 
The circle with the minor axis as diameter is called the 
minor auxiliary circle or the minor circle. 

If B be any point on the major auxiliary circle, then 
when this circle is turned about AA' and projected into 



IV 



CONIC SECTIONS 



79 





T '' i 


%C ' s 


B 


o/\\ 






< GL^^Ti 








S / 


r\ M\ 






\ 




^fT^V 


1 (a: \ 




A\\ 


\ 


R 


b lA 


^ 


C 
B' 


\M 


\A 



the ellipse, the projection of P Q will be P, where P P is 
perpendicular to CA. And from the principles of projection 

CA : CB = constant 



P M: PM=P C: BC-- 



for all points on the curve. 

Again, if the ellipse be turned about the minor axis 
and projected into the minor auxiliary circle, the point P 
projects into P x , where PP 1 is perpendicular to CB and 



PN\ P X N= AC: A 1 C=CA: CB 



P M: PM. 



From this relation it is easy to prove that the line P P 1 
if produced will pass through C. 

These theorems may be stated as follows : 

Theorem i. The ratio of the ordinate of any point of 
an ellipse to the corresponding ordinate of the major auxiliary 
circle, is constant and equal to the ratio of the mi?ior to the 
major axis. And this is equal to the inverse ratio of the 
abscissa of the point to the abscissa of the correspo?iding point 
on the minor auxiliary circle. 

See Art. 144 for definitions of ordinate and abscissa. 



So PRACTICAL PLANE GEOMETRY chap. 

Theorem 2. If P be any point on an ellipse, ike 
corresponding points P and P x on the auxiliary circles tie on 
the same radius. 

Next, let the tangent at any point Q on the ellipse 
intersect the major and minor axes in R and S. If 
the ellipse, with the tangent, be turned about the minor 
axis so as to project into the minor auxiliary circle, then Q 
will come to Q v S will not move, and the tangent SQ to 
the ellipse will project into the tangent SQ 1 to the minor 
auxiliary circle. Similarly RQ will be the tangent to the 
major auxiliary circle corresponding to tangent RQ to the 
ellipse. And RQ , SQ l are parallel to each other since 
<2 , <2i ue on tne same radius ; we thus have the following 
theorem : 

Theorem 3. A tangent to an ellipse and the corre- 
sponding tangent to the auxiliary circle intersects the major 
axis at the same point. Also the tangent to the ellipse and 
the corresponding tangent to the minor auxiliary circle 
intersects the minor axis at the same point. And the two 
auxiliary tangents are parallel to each other. 

Bisect P X P in O and join OP. Then it is clear that 
for all positions of the point P 

nn CP, + CP n CA + CB 

CO = - 1 = = constant. 

2 a 

and also that 

CA - CB 

OP = OP 1 = OP = = constant. 

Further, since both the triangles PfDP, P Q OP are 
isosceles, it is evident that CO and OP are always equally 
and symmetrically inclined to CA ; they are also equally 
and symmetrically inclined to CB. We are thus led to the 
following theorem : 

Theorem 4. If two lines CO and OP of constant length 
move round a fixed point C so as to be always symmetrically 
inclined to a given fixed line, the locus of P will be an ellipse 



IV 



CONIC SECTIONS 




of which the principal axes are lines through C parallel and 
perpendicular to the fixed line, the lengths of the semi-axes 
being respectively equal to the sum and difference of CO and 
OP. 

A mechanism for draining ellipses has been constructed 
on this principle. CO is a crank free to rotate on a pin 
fixed at C to a frame. OP is a second crank pinned 
to the first one at O, and constrained by mechanism to 
turn (i.e. alter its angular position) at the same rate that 
CO turns, but in the opposite direction. In order to be 
able to trace an ellipse of any given size, it is necessary that 
the lengths of both cranks be capable of adjustment, one 
being made equal to half the sum, and the other to half 
the difference of the semi-axes. The figure shows OP as, 
the shorter crank, but the same ellipse would be traced if 
the lengths of CO and OP were interchanged so that OP 
were the longer. 

Example. Construct the above figure when the lengths of the 
axes AA', BB' are 4" and 2j". 

G 



82 PRACTICAL PLANE GEOMETRY chap. 

83. Problem. To construct an ellipse, having given 
the major and minor axes. (Second method.) 

Let A A' and BB' be the major and minor axes. 

With centre C describe the two auxiliary circles on A A ' , 
BB' as. diameters. 

Divide the arc AB into a number of equal parts, 
say six ; draw the radii to the points of division, meeting 
the inner circle in corresponding points. Draw lines 
through the outer set of points perpendicular to the major 
axes to meet parallels thereto through the inner set. 

A fair curve must be drawn through the points thus 
found. The remaining portions of the ellipse may be 
similarly determined, or obtained by constructions depend- 
ing on symmetry. 

84. Problem. Having given the principal axes of an 
ellipse, to find points in the curve mechanically by means 
of a paper trammel. 

Let AA', BB' be the two axes. Take a strip of paper 
with one edge straight, and from a point P on this edge 
mark off Pa, Pb, equal to CA, CB, the semi-axes. 

Place the strip in successive positions with a and b 
always on the minor and major axes respectively, and mark 
corresponding positions oi P \ these will be points on the 
ellipse. 

For, bisect ab in O and join CO. Then 
CO = Ob = \ab = constant 
and angle OCb = angle ObC. 

Hence CO, OP, two lines of constant length, move round 
C, and in all positions are symmetrically inclined to a fixed 
line CA ; therefore by Theorem 4, Art. 82, the locus of P 
is an ellipse of which 

the semi-major axis = PO + CO = PO + Oa 
= Pa = CA 
and the semi-minor axis = PO - CO = PO - Ob 

= Pb = CB. 
An instrument called a trammel for drawing ellipses is 
constructed on this principle. 



IV 



CONIC SECTIONS 



3 




If Pa and Pb had been set off opposite ways along the 

edge of the trammel, the same ellipse would have been 

described. CO would then have been the longer of the 

two cranks. 

Xotc. Tracing-paper may with advantage be used for the trammel, 
in which case the points /*, a, b need not be on its edge. 



84 PRACTICAL PLANE GEOMETRY chap. 

85. Problem. A triangle moves in such a manner 
that two of its angular points always lie respectively in 
two fixed lines, to determine the locus of the third 
angular point. 

Let PQR be one position of the triangle, the points Q 
and R being on the fixed lines CVand CX respectively ; 
the locus of P is required. 

Describe a circle to pass through C, Q, R, and let O be 
its centre; join and produce PO, cutting the circle in b and a. 
Draw the lines CaB, CbA, and make CA = Pa and CB 
Pb. The locus of P will be an ellipse of which CA and 
CB are the major and minor semi-axes. 

For, join OR, and suppose the circle and the lines 
PO, OR to be drawn on the plane of the triangle and to 
move with it. 

The radius of the circle, the chord QR, and the angle 
QCR remain constant in magnitude, therefore as the triangle 
and circle move together the latter always passes through 
C (Euc. III. 21). Hence 

CO moves about C as centre . . . (i.) 
Now suppose that the triangle turns through any angle 0, 
say clockwise; then OR will do the same (since it is 
attached to the triangle), and the angle ORC will be 
increased by an amount 6. But since OCR is isosceles 
the angle OCR will also be increased by an amount 6, and 
therefore OC will turn through an angle anti-clockwise. 
Hence 

CO and OP turn about C and O respectively 
at the same rate but in opposite directions . (ii.) 
But in the position shown in the figure, CO and OP are 
symmetrically inclined to the fixed line CA, since OCb is 
isosceles, therefore from (ii.) 

CO and OP turn about C and are always 
symmetrically inclined to the fixed line CA (iii.) 

Hence from (iii.), as stated in Theorem 4, Art. 82, it 
follows that the locus of P is an ellipse of which 



IV 



CONIC SECTIONS 



35 




c 



the semi-major axis = PO + OC '= Pa = CA 
and the semi-minor axis = PO - OC Pb = CB, 

the directions of which are CA and CB. 

It will be noticed that Pba is the trammel for the double 
crank CO, OP. 

The triangular trammel PQP, the double crank CO, OP, 
and the principal straight line trammel Pba are thus seen to 
be equivalent to one another in determining the path of P. 

An indefinite number of equivalent trammels could be 
found, either triangular or straight ; for the former by taking 
any two fixed lines CX\ CY', intersecting the circle in 
R', Q', and joining PQ', PR' ; or for the latter by drawing 
any line through P intersecting the circle in a, b', and then 
taking Ca', Cb' as the fixed lines. 



Examples. 1. The lengths of the major and minor axes of an 
ellipse are respectively 4" and 3". Set out the curve (a) by the 
method of Prob. 83 ; (6) by the method of Prob. 84, using 
tracing-paper for the trammel. 

2. Work Prob. 85, having given 

Angle XCY=-jo, QP=2.$", QP=2.-j$", RP= 1.0". 
Measure the lengths of the principal axes of the ellipse described 
by P, and the angle which the major axis makes with CA'. 
Ans. 6.55", 1.27", 13. 7 . 



86 PRACTICAL PLANE GEOMETRY chap. 

86. Conjugate diameters of an ellipse. 

Let tangents be drawn to a circle at the ends of two 
diameters perpendicular to each other, thus forming a 
circumscribing square as shown in the figure ; and let PQ, 
PR be any two chords of the circle respectively parallel 
to the diameters. 

Suppose this figure to be turned about any line XX 
so that its plane is inclined to the plane of the paper, 
and the northogonally projected ; the shape of the projec- 
tion will be as shown to the right, corresponding points 
being denoted by corresponding letters. 

From the principles of orthogonal projection we infer 
that the square with the inscribed circle touching it at the 
middle points of its sides will project into a parallelogram 
with an inscribed ellipse touching the sides at their middle 
points. It also follows that </and kg, the tangents at the 
ends of the diameter jf, are parallel to the diameter i'i ; 
similarly the tangents eh and fg at the ends of the diameter 
i'i are parallel to the diameter jf. 

And since the chords PQ and PR are bisected by JJ' 
and 77' at M and irrespectively, so also are the chords/^ 
and pr bisected by jj' and ii' at m and n respectively. 

Such diameters of an ellipse as ii' and^/' are said to be 
conjugate ; the tangents to the ellipse at the ends of either 
diameter are parallel to the other diameter ; each diameter 
bisects all chords parallel to the other. 

87. Problem. Having given a pair of conjugate 
diameters of an ellipse, to determine the principal axes. 

Let 77', JJ' be the given conjugate diameters. Draw 
JG perpendicular to and equal to CI. Join CG, and on 
CG as diameter describe a circle, centre O. Join JO, 
intersecting the circle in b and produced in a. Join Cb, 
Ca, and on these lines produced take CA and CA', each 
equal to Ja, and CB, CB ', each equal to Jb. 

Then AA' and BB' are the major and minor axes of 
the ellipse. 



IV 



CONIC SECTIONS 



87 




For, join if, then by comparing with Prob. 85 it is seen 
that Jij is the triangular trammel which will give the re- 
quired ellipse ; CO, OJ is the double crank, and Jba the 
principal straight line trammel equivalent to the triangular 
trammel in Prob. 85. 

Example. The lengths of two semi-conjugate diameters of an 
ellipse are 4" and 3", and the included angle 6o 3 ; determine 
the principal semi-axes, and the angle which the major axis 
makes with the larger conjugate diameter. 
A ns. 4.40", 2.35", 1 7. 5 , 



88 PRACTICAL PLANE GEOMETRY chap. 



88. Problem. To describe an ellipse, having given a 
pair of conjugate diameters, or to inscribe the principal 
ellipse in a given parallelogram. 

Let DEFG be the given parallelogram, or II', Jf the 
given conjugate diameters. 

First Method. On two adjacent sides of the parallelo- 
gram, DG, DE, as diameters, describe semicircles, and 
divide each into the same number of equal parts, say six. 

From the points of division draw lines perpendicular to 
the sides, intersecting the latter in points marked i, 2. 

From the points on DG draw lines parallel to DE ; and 
from the points on DE draw lines parallel to DG. 

These two series of lines will intersect in a series of 
points i p 2j on the required ellipse. 

Second Method. Divide CI and EI each into the same 
number of equal parts, say three, and draw lines from /' 
and /to the points 1, 2, as shown. 

The intersections of corresponding lines will give points 
1 2 j, which are on the required ellipse. 

Proof. The above two constructions are readily seen to be true for 
a square with inscribed circle, of which the figure may be regarded 
as the projection. Hence they are true for the projection. 

Third Method. By a triangular trammel. 

Draw JjG x perpendicular to CI, and make JG 1 = CI. 
Draw / perpendicular to Jf and join ij. Then Jji is the 
shape of the required triangular trammel. 

Take a piece of tracing-paper for the trammel and mark 
the three points, /, /, / on it ; then if i and / move on JJ' 
and W respectively, P (or /) will trace the ellipse. 

It will be noticed that the angle ijj (or iPj) is equal to 
the complement of the angle ICJ, and that the lengths of 
Pi and Pj are respectively equal to the perpendicular 
distances of /and /from the conjugate diameters. 

Proof. The point G x is the instantaneous centre of rotation of 
the trammel when moving through the position shown ; hence the 
ellipse described by P touches DE at/. It can be shown in a similar 
manner that the ellipse touches EF sX I. 



IV 



CONIC SECTK )NS 



89 



/'..- 



88 




89. Problem. Having given the curve of an ellipse, 
to determine the principal axes. (No figure.) 

First Method. Draw any two chords, PP', QQ', parallel 
to each other. Draw the line bisecting these chords and 
meeting the curve in /, I', then IP is a diameter, and C, 
its middle point, is the centre of the ellipse. 

Through C draw the diameter JJ' parallel to PP' . 

Two conjugate diameters are thus determined, and the 
major and minor axes can be found as in Prob. 87. 

Second Method. Proceed as in the first method so far 
as to find C. 

Then with centre C and any suitable radius, describe a 
circle cutting the ellipse in P, Q, P, S (taken in order). 

Draw the diameters PP, QS, and bisect the angles 
between them. The bisecting lines terminated by the 
curve are the principal axes. 

Examples. 1. Two conjugate diameters of an ellipse are 4-^" and 
3V' long and make 6o with one another ; describe the ellipse 
by each of the three methods of Prob. 88. 

2. Rule any two intersecting lines in ink on drawing-paper. Mark 
any three points on thick tracing-paper. Let two of the points 
move one on each line, and prick off points on the ellipse traced 
by the third point. Draw a fair curve through the points. 
Find the principal axes of the ellipse by each of the methods of 
1'rob. 89. Confirm the result by the construction of Prob. 87. 



9 o PRACTICAL PLANE GEOMETRY chap. 

90. Tangent and normal to any plane curve. Curvature. 

Let P, Q be two points in any given curve, and 
suppose the point Q to move on the curve and to approach 
indefinitely near to P, then the chord PQ will turn round 
P and tend to a definite direction PT, becoming in the 
limit the tangent to the curve at P. The line PG, per- 
pendicular to PT, is called the normal to the curve at P. 

Definitioti i. The tangent at a point in a curve is the 
straight tine passing through the point and through a second 
point in the curve indefinitely near the first. The normal to 
the curve at the point is the straight line passing through the 
point at right angles to the tangent. 

This general definition of a tangent does not indicate a practical 
method of accurately drawing it at any given point in the curve. In 
order to be able to draw the tangent in the right direction it would be 
necessary to know something more with regard to it, such as the 
position of a second point in it at a finite and sensible distance from 
the first point (or a line to which it is parallel, perpendicular, or 
inclined at a known angle, or a second curve which it touches, etc.), 
and this would require special knowledge of the particular curve under 
consideration. 

If it be required to draw a tangent from a point R not in the given 
curve, this may be done with all attainable precision, by applying a 
straight-edge, adjusting its position and then drawing the line RS to 
touch the curve as shown. The exact point of contact at S would be 
still undetermined, and if this point, as well as the line of the tangent, 
were required, an additional construction would be necessary, such as 
the drawing of a line intersecting the tangent at the required point. 
This additional construction would require to be based on some known 
property of the given curve. 

Let Q, P, R be any three points on a plane curve ; 
bisect the two chords QP, PR at right angles by the lines 
AC, BC meeting at C ; then a circle with centre C passing 
through P will also pass through Q and R. Suppose now 
the two points Q and R to move along the curve and to 
approach indefinitely near to P. Then in the limit the two 
bisectors AC, BC become normals to the curve; they 
intersect in a definite point O, called the centre of curvature. 
The circle through P with centre O is called the circle of 
curvature at P, and OP is the radius of curvature. 



IV 



CONIC SECTIONS 



91 




Definition 2. The circle of curvature at any point in a 
curve is the circle passing through the point and through two 
other points in the curve indefinitely near the first. Its radius 
is the radius of curvature, and its centre, called the centre of 
curvature, is the point of intersection of the normal at the 
point on the curve, with a second normal at a point indefinitely 
near the first. 

The centre and radius of curvature cannot be accurately determined 
by directly applying the above construction ; a special construction is 
required, depending in each case on the nature of the curve. 

Any circle through P with its centre on the normal at P would 
touch the curve, the two would have a common tangent, or have two 
consecutive points in common ; but the circle of curvature coincides 
more closely with the curve at the point than any other circle which 
can be drawn ; as we have seen, it has three consecutive points in 
common with the curve. On examining the figure it will be noticed 
that the circle of curvature crosses the curve at P ; on one side of P 
the curve gradually becomes flatter, and so falls away from the circle 
on the outside ; on the other side of Pihe. curvature gradually increases, 
and so the curve bends away from the circle towards the inside. It 
is thus seen that the amount of curvature at P is correctly measured 
by the circle of curvature ; and the latter always crosses the curve 
unless the curvature is a maximum or minimum, as, for example, at 
the ends; of the principal axes of an ellipse ; in this case the curve and 
the circle of curvature havefo/tr consecutive points in common. 



92 PRACTICAL PLANE GEOMETRY chap. 

91. Focal and tangential properties of the ellipse. 

Let PT, PG be the tangent and normal to the ellipse 
at any point P on the curve. 

Draw FH, F H' perpendicular to the tangent. Draw 
the focal lines FP, FP, and produce one of them. 

Then it may be shown that the angle FPG = angle 
FPG, and that the angle FPT= angle APT. Also that 

ar=cn'=CA. Or 

Theorem i. The normal at any point P in an ellipse 
bisects the ang/e between the focal lines PF, PF, and the 
tangent bisects the angle between one of the focal lines and 
the other one produced. 

Theorem 2. The feet of the perpendiculars from the 
foci to any tangent to an ellipse lie on the major auxiliary 
circle. 

92. Problem. To draw the tangent and normal to a 
given ellipse at a given point on the curve. 

First Method. Let P be the given point. Join FP, 
F'P; produce one of these, say F'P to A', and bisect the 
angle FPA'by the line PT; then PT is the tangent at P. 

Draw PG perpendicular to PT, or bisect the angle 
FPF; then PG is the normal at P. 

Second Afethod (refer to figure on p. 95).- Let Q be 
the given point. Determine Q , the corresponding point 
on the major auxiliary circle. Draw the tangent Q P, 
intersecting the major axis in P ; join RQ ; then PQ is the 
required tangent. 

Or if P be too remote, determine Q v draw the tangent 
Q-yS; then SQ is the required tangent. 

The normal at Q is not shown in the figure, but it 
passes through Q, and is perpendicular to PS. 

Third Method, (see Fig. 84). Let P be the given point. 
With centre P and radius equal to CA, describe an arc 
intersecting the minor axis in a. Join Pa, intersecting the 
major axis in b. Through a and b draw lines (not shown) 
perpendicular to axes BB ', A A', intersecting in G ; then 
PG is the required normal. 



IV 



CONIC SECTIONS 



93 




For G is the instantaneous centre of the trammel Pba, and therefore 
the tracing-point P is moving in the direction at right angles to PG. 

93. Problem. Having drawn a tangent to a given 
ellipse, to find the point of contact. 

First Method. Let the tangent be the one shown in the 
figure above. From one focus T^draw FHK perpendicular 
to the tangent intersecting it in H, and make HK=FH. 
Join KF' intersecting the tangent in F ; then P is the 
point of contact. 

Second MetJiod (see the figure on p. 95). -Let the tangent 
intersect the major and minor axes in R and S. 

From R draw a tangent RQ to the major auxiliary circle, 
and determine the point of contact Q by drawing CQ 
perpendicular to RQ - Draw QQ perpendicular to A A' 
to meet the tangent in Q. Then Q is the required point 
of contact. 

If the point R is not available, make the corresponding 
construction for the minor auxiliary circle, i.e. draw SQ V 

CQ V Q X Q- 

Example. Draw the above figure one-half larger. 



94 



PRACTICAL PLANE GEOMETRY chap. 



94. Problem. Having given the principal axes (or the 
foci and one axis) of an ellipse, to determine the two 
tangents to the curve from a given external point (with- 
out drawing the curve), and to find the points of contact. 
First Method (figure, last page). Let T be the given 
point. Join T to one of the foci, say F. On TF as 
diameter describe a circle, and draw also the auxiliary 
circle on AA'. Let these circles intersect in B, H x ; 
then TJ7, TH X are the required tangents. 

To find the points of contact, produce FH to K, and 
make HK=HF. Join KF, intersecting the tangent in 
F; then F is the required point of contact for one of the 
tangents ; that for the other is found in a similar manner. 

Second Method (figure opposite). This method is based 
on the projective properties relating to the ellipse and the 
auxiliary circles. See Art. 82. 

First let the given point be R, on the major axis. 
Draw RQ to touch the major circle, and draw CQ 
perpendicular to FQ , thus determining the exact point of 
contact <2 , and also the point Q x on the minor circle. 

Through Q Q , Q x draw the lines parallel to the axes, meet- 
ing in Q. Then RQ is a tangent, and Q the point of contact. 
Next let the given point be S, on the minor axis. 
Draw SQ V tangent to the minor circle, then draw the 
perpendicular CQ X Q^ thus determining Q v Q , from which 
Q may be found as in the last case. 

Lastly, let the given point be T, on neither axis. 
Draw TL perpendicular to AA'. Make LD = LT. 
Draw DT parallel to A X 'B (or perpendicular to AB\ 
intersecting LT produced in T v Then T is a point in 
the auxiliary tangent to the major circle. 

For 7' Z : TZ = B C : FC= Q F : QE. 
This tangent T Q is then drawn, and the points Q , Q v Q 
found as before. The required tangent from T is TQ, and 
Q is the point of contact. 

The second tangent would correspond with the second 
tangent (not shown) from T Q to the major circle. 



IV 



CONIC SECTIONS 



95 




Examples. 1. Draw the above figure double size. 

2. Using tracing-paper for a trammel, construct an ellipse whose 

principal axes are 4^" and 3" long. Select any point P on the 
curve, and determine the tangent and normal at P by each of 
the methods of Prob. 92. 

3. In Ex. 2 select any point Q external to the ellipse, and through 

Q draw a tangent to the curve. Then determine the point of 
contact by each of the methods of Prob. 93. 

4. Draw two lines AA ', BB ', 4" and 3" long, bisecting each other 

at right angles in C. Along CA, CB produced, set off CR, 
CS equal to 3.6" and 2.8"; and mark a point T such that 
AT=.f, BT=2.7". 

By each of the methods of Prob. 94 draw from R, S, and T 
tangents to the ellipse of which AA', BB' are the principal axes, 
without drawing the curve, and find the points of contact. 

5. Find the eccentricity of the ellipse of Ex. 2, and the distance 

apart of the directrices. Ans. 0.745: 6.04". 
Hint. See Theorem 2, Art. 77. 

6. Having given one focus F and two tangents TP, TP' with their 

points of contact P, P' ; to find the other focus F' . 

Hint. See Fig. 91. Draw FHK and join KP. Similarly 
obtain KP' . Then F' is at the point where KP and K ' P' 
meet. 



96 PRACTICAL PLANE GEOMETRY chap. 



95. Problem. Having given the major and minor 
axes of an ellipse, to find, without drawing the ellipse, 
the points (if any) in which a given line intersects the 
curve. 

The method adopted is based on the projective pro- 
perties of the auxiliary circles described in Art. 82. 

Draw the auxiliary circles on the given axes AA' and 
BB' as diameters. Let the given line intersect the major 
auxiliary circle in L, M. 

Join CL, CM, cutting the minor auxiliary circle in /, m. 
Through L, M draw lines parallel to A A' to meet in L v M 1 
lines through /, /// drawn parallel to BB'. 

Join L X M X intersecting the minor auxiliary circle in P v 
Q v Through P v Q x draw lines parallel to AA' to meet 
LM in P, Q\ then P and Q are the required points of 
intersection of LM and the ellipse. 

If the given line be perpendicular to the major axis, let 
H be the point in which it meets the major circle ; join 
CH, cutting the minor circle in D , and draw DAI perpen- 
dicular to BB' to meet the given line in D ; then D is one 
of the required points of intersection. 

If the given line be perpendicular to the minor axis, let 
K be the point in which it meets the minor circle. Join 
CK and produce it to meet the major circle in P ; draw 
E Q E perpendicular to AA' to meet the given line in E ; 
then E is one of the required points of intersection. 

96. Problem. To describe an ellipse having given its 
centre C and three points P, Q, R on the curve. 

Join CQ and. PR intersecting in H. With centre C 
radius CQ, describe a circle, and through Zf draw the chord 
pr of this circle, such that/ZT: Hr = PH\ HR ; and draw 
the radius Cs perpendicular to CQ. Join sr (or sp) inter- 
secting CQ in A". Join and produce RK (or PR) to meet 
in S a line drawn through s parallel to Rr or Pp, and join 
CS. 

Then CS, CQ, are conjugate semi - diameters of the 



IV 



CONIC SECTIONS 



97 



96 




SyS. 



required ellipse. The principal axes can be determined 
by Prob. 87. 

This construction is based on some theorems of projection. The 
required ellipse is supposed to be projected into a circle. A figure 
similar to the projection is drawn of such a size and so placed that CQ 
coincides with its own projection. The first part of the construction 
locates / and r, the projections of P and R. The second part locates 
S, a point which projects into s, where Cs has been drawn conjugate 
to CQ in the projection. 

H 



98 PRACTICAL PLANE GEOMETRY chap. 



Examples. 1. In Ex. 4, p. 95, join RT. Also draw a line DD 
parallel to A A' at a distance of .5" ; and a line EE parallel to BB' 
and distant .7" therefrom. By the method of Prob. 95 determine the 
points in which the lines RT, DD, EE would cut the ellipse, if the 
latter were drawn. 

2. Work Prob. 96, taking for the data CP=l.$" i CQ=l.f, CR 
= 2.2"; PCQ=72, QCR=34 a . Complete the figure by drawing 
the ellipse through P, Q, R. 

97. Problem. To construct an ellipse, having given 
the foci F, F' and a tangent. 

(See figure, Art. 91.) From one of the given foci, F, 
draw a perpendicular to the given tangent meeting it in H; 
then H is on the major auxiliary circle. 

Bisect FF' in C, and with centre C and radius CH 
describe the circle cutting FF' produced in A, A' ; then 
AA' is the major axis. 

The minor axis may now be found as in Prob. 79, and 
the ellipse constructed by one of the methods already given. 

98. Problem. Having given one axis of an ellipse 
and a tangent, to construct the curve. 

If the major axis be given, describe the auxiliary circle 
on it, and from the two points in which the given tangent 
cuts the circle, draw lines perpendicular to the tangent, to 
meet the major axis in F, F', the foci (see figure, Art. 91). 
The minor axis can now be found and the ellipse con- 
structed. 

If the minor axis be given, describe the minor circle on 
it and let .S (figure, page 95) be the point in which the 
given tangent cuts the given axis (produced if necessary). 

Draw the tangent from S to the circle, and from C draw 
CQ X perpendicular to the tangent. 

Draw Q x Q perpendicular to BB' to meet the given 
tangent in Q, and from Q draw QQ parallel to BB' to 
meet CQ X produced in Q Q ; then Q is a point on the 
major auxiliary circle, and hence the major axis is known 
and the ellipse can be constructed. 



IV 



CONIC SECTIONS 



99 



THE PARABOLA 

99. Problem. To construct a parabola, having given 
the focus F and directrix DD'. 

It has been explained in Art. 75 that the section of a cone by a 
plane parallel to a tangent plane is a parabola. And in Art. 76 it was 
shown that this curve may also be denned as the locus of a point 
which moves in a plane in such a manner that its distance from a 
fixed point, the focus, is always equal to its distance from a fixed line, 
the directrix ; this property enables us to set out the curve. 

Through i^draw a line perpendicular to DD\ meeting 
the latter in Z. Bisect ZF in A. 

Then A is on the parabola, for AF= AZ. A is called 
the vertex, and the perpendicular ZF the axis of the para- 
bola. The latter is a line of symmetry for the curve. 

To draw the curve. Select a series of points 1, 2, 3 . . . 
on the axis as shown, and through these points draw lines 
perpendicular to the axis. 

With F as centre, Z\ as radius, describe an arc to inter 
sect the perpendicular through 1 in i r 

Repeat this construction for the other points, and draw 
a fair curve through the points A, i p 2 p 3 X . . . thus 
found. 




ioo PRACTICAL PLANE GEOMETRY chap. 

100. Properties of the parabola. In the upper figure 
PN and PM are perpendicular to the axis and directrix. 
PT is the tangent and PG the normal at P, and A Y is the 
tangent at the vertex. PN is called the ordinate, and NG 
the subnormal for the point P. The double ordinate LL X 
through the focus is called the latus rectum. 

In the lower figure let PQ be any chord of a parabola; 
PT, QT the tangents at P and Q ; TN a line parallel to 
the axis, intersecting the curve at A ; and let XY be the 
tangent at A. 

The following theorems are proved in works on pure 
mathematics. The student should commit them to 
memory, and test them by setting out the figures to scale. 

Theorem i. The tangent PT bisects the angle PPM. 

Theorem 2. A line drawn through the focus perpendicular 
to a tangent, meets the latter at a point which is on the tangent 
at the vertex. 

Theorem 3. AT and AN are equal. (In both figures.) 

Theorem 4. The latus rectum LL X is equal to 4AP. 

Theorem 5. The length of the subnormal NG is constant, 
and equal to half the latus rectum, or to 2AP, or to PZ. 

Theorem 6. PN 2 is proportional to AN. Or the square 
of the ordinate is proportional to the abscissa. For the upper 
figure PN 2 = Z l -AN=4AF-AN Also for the lower 
figure PN 2 ^AF- AN, where Pis the focus. 

Theorem 7. A line which bisects any system of parallel 
chords is called a diameter, and is parallel to the axis, which 
is a particular diameter. 

Theorem 8. The line TN, through T, parallel to the axis, 
bisects the chord PQ, and TN is bisected at A. That is, PN 
= NQ, and TA = AN 

Theorem 9. The tangent XY at A is parallel to PQ, and 
XY= \PQ. 

Theorem 10. Tangents which meet in the directrix are at 
tight angles to each other. 

Theorem 1 1. The area ALPNA is two-thirds the circum- 
scribing rectangle AN- PN. 



IV 



CONIC SECTIONS 



IOI 




101. Problem. To draw the tangent and normal 
to a given parabola at a given point P on the curve. 

(See the upper figure.) Join P to the focus F, and draw 
PM parallel to the axis ; bisect the angle PPM by the 
line PP. Then PP is the tangent at P. 

Or, make PP= PP and join PP. 

Or, make AP=AJV and join PP. 

To draw the normal, first draw the tangent as above, 
and then draw the normal PG perpendicular to it. 

Or, make NG = PZ, and join PG. 



io2 PRACTICAL PLANE GEOMETRY chap. 



102. Problem. To draw the tangent to a parabola 
from an external point T. 

Join T to the focus F, and on TF as diameter describe 
a circle to cut the tangent at the vertex in H and H' . 
Then FH, TH' are tangents to the parabola. 

103. Problem. To describe a parabola, having given 
two tangents with their points of contact. 

Let FT, QT be the given tangents, intersecting in T, 
and let F, Q be the given points of contact. 

Join FQ and bisect FQ in N Join FN and bisect 
FN in A. Draw through A a line Rt parallel to QF. 

Then 7Wis parallel to the axis of the parabola, A is a 
point on the curve, and FA is a tangent at A. 

First Method. Shown on the left of FN. Draw QR 
parallel to NA, so that QRANis a parallelogram. 

Divide NQ and RQ into any and the same number of 
equal parts, say four. From the points of division i, 2, 3 
on NQ draw lines parallel to NA ; and from the points of 
division 1, 2, 3 on RQ draw lines through A. 

Let the parallel lines intersect the corresponding radial 
lines in the points I. II. III., as shown. A fair curve 
through these latter points is the parabola required. 

The curve may be extended beyond Q by continuing 
the divisions 5, 6, ... on NQ, RQ, and through them 
drawing the parallels and radials, intersecting in V. VI. . . . 

The same construction might be used for the portion of 
the curve AF. Instead we give the following : 

Second Method. Shown on the right of FN Deter- 
mining A as before. Let the tangent at A, which is 
parallel to QF, intersect FF'm t. 

Join FA, and through / draw a line parallel to AN, 
intersecting FA in n. Bisect hi in a. Then a is a point 
on the curve, and the tangent at a is parallel to FA. 

This construction for determining a is the same as the 
one that was used for finding A. If desired, further points 
between Aa and aP can be found by repeating the process. 



IV 



CONIC SECTIONS 



103 




104. Examples on the parabola. 

1. Construct a parabola for which the distance of the focus from 

the directrix is ". Measure the latus rectum. Ans. ij". 

2. In Ex. 1 take a point P on the curve distant 2" from the focus, 

and draw the ordinate, the tangent, and the normal at /'. 
Measure the length of the subnormal. Ans. ^". 

3. In Ex. 1 select any point external to the parabola, and draw a 

tangent to the curve. Determine the point of contact. 

4. Draw a triangle TPQ, having given PQ-3", PT= 2f", QT= 2". 

By each of the methods of Prob. 103 construct the parabola 
which touches PT and QT at /'and Q. 

5. Determine the axis, vertex, focus, and latus rectum of the para- 

bola of Ex. 4. Ans. Length of the latus rectum = 2.24". 

6. Draw any parallelogram, and in it inscribe a parabola which 

touches one side at its middle point, and passes through the 
ends of the opposite side. Determine the latus rectum. 

7. A jet of water issuing from a horizontal nozzle strikes a point I 

foot below the orifice and 3 feet distant horizontally. Draw 
the path of the jet ; scale x \. Find the latus rectum of the 
path, and obtain the horizontal velocity of the water. Ans. 9 
feet ; 12 feet per second. 

Hint. It is known that (neglecting air resistances) the path 
of an object moving freely under the action of gravity is a para- 
bola, with axis vertical and latus rectum in feet equal to z> 2 -r 16, 
where v is the horizontal velocity in feet per second. 



104 PRACTICAL PLANE GEOMETRY chap. 



THE HYPERBOLA 

105. Properties of the curve. "\Ye explained in Art. 
75 that a hyperbola was obtained when a cone was cut by 
a plane so taken as to penetrate both portions of the com- 
plete surface ; and in Art. 76 it was shown that the curve 
might be defined as the locus of a point which moves, so 
that its distance from a fixed point bears a constant ratio 
(greater than unity) to its distance from a fixed line. 

The curve is set out to scale in the figure ; it is seen to 
consist of two parts detached from each other, each part 
having two branches ; these branches extend to infinity on 
the complete curve. 

Although so different in appearance from the ellipse, 
the two curves have many closely allied relations. Both 
have two axes of symmetry, two foci, two directrices, and 
each has a centre. Each curve, however, has properties 
peculiarly its own. 

The hyperbola is not treated as fully in this work as is 
the ellipse, on account of its minor importance in the arts. 
We shall define the terms, state some of the properties, 
and give the more useful problems. The student should 
illustrate the subject by drawings to scale. Proofs may be 
found in works on pure mathematics. 

Definitions. The points F, F' are the foci. 

C is the centre, chords through which are bisected at C. 

A and A' are the vertices, and AA' is the transverse axis. 

BB ', perpendicular to AA', is called the conjugate axis, 
being determined by the condition that AB = AB' = CE 
= CE' = CF. 

The diagonals DE', D ' E, produced indefinitely both 
ways, are asymptotes to the curve. 

Asymptotes are lines which, as they recede to infinity with 
a curve, approach nearer a?id nearer to the latter without limit, 
but never actually coincide with it. 

Properties. Let P be any point on the curve. [oin P 
to Fznd F' ; then P'F- PF= A A' ; or 



IV 



CONIC SECTIONS 



105 




Theorem 1. The difference of the focal radii is constant, 
and equal to the transverse axis. 

Observe that in the ellipse it is the sum which is constant. 

Theorem 2. The tangent PT at any point P bisects the 
focal radii PP, PP' . 

In the ellipse it is the normal which bisects this angle. 

From any point Q on the curve, draw lines parallel to 
the asymptotes, meeting the latter in M or N. Then 

Theorem 3. The product QM QN is constant. 
This important property is characteristic of the hyperbola. 

Let X be the point where a directrix cuts CA (not 
shown), then it may be shown that 

Theorem 4. The eccentricity FA : AX CA : CX 
= CP: CA, and CX, CA, CF are in geometrical progression. 

Compare Theorem 2, Art. 77, for the ellipse. 



io6 PRACTICAL PLANE GEOMETRV chap. 



106. Problem. To construct a hyperbola, having given 
the focii F, F', and the transverse axis AA' (Fig. 105). 

Select points 1, 2, 3, 4 . . . on the axis, outside the 
foci, say to the left of F 1 . 

With radius Ai, centres F and F, describe arcs ; and 
with radius A'i, centres F' and F, describe arcs intersect- 
ing the first arcs in the points marked 1'. 

Repeat this construction for the points 2, 3, 4 . . . 
obtaining the points marked 2, 3', 4 . . . A fair curve 
drawn through the latter points is the hyperbola required. 
This construction is based on Theorem 1, Art. 105. 

107. Problem. Having given the foci F, F' and trans- 
verse axis AA' of a hyperbola, to determine the asymptotes 
DE', D'E, and the conjugate axis BB' (Fig. 105). 

Bisect A A' in C. With centre Cand radius CF describe 
arcs (only half of one shown) intersecting the perpen- 
diculars to the axis through A and A' in F, F', >, )'. 

Then the lines, of indefinite length, through D', E and 
D, F' are the asymptotes. 

To obtain the conjugate axis draw a perpendicular 
through C, and cut this perpendicular in B, B' by an arc 
drawn with A as centre, CF as radius. Then BB' is the 
conjugate axis. 

Or make CB and CB' each equal to AF. 

Note that in the ellipse FB- CA, and in the hyperbola AB-CK 

108. Problem. To draw the tangent and normal to a 
hyperbola at any point P on the curve (Fig. 105). 

Join PF, PF' and bisect the angle FPF' by the line 
PT, then PT is the tangent at P. 

Draw PG perpendicular to PT, then PG is the normal. 

Example. The transverse axis of a hyperbola is 2j" and the 
distance between the foci is 2-|". Determine the conjugate 
axis, the asymptotes, and draw a portion of the curve. 

Draw the tangent at any point and illustrate that this point 
bisects that part of the tangent between the asymptotes. 

Find the eccentricity and the distance apart of the directrices. 
Ans. 1.58", 1.22, 1. 84". 



IV 



CONIC SECTIONS 



107 




109. Problem. To construct a hyperbola, having given 
the asymptotes XX', YY', and a point P on the curve. 

Through the given point P draw lines PAf, PN of 
indefinite length, parallel to the asymptotes. 

Through the point O, where the asymptotes intersect, 
draw a series of radials cutting the lines PM, PN in the 
points 1, 1 ; 2, 2 ; 3, 3; 4. 4. 

From the points 1 and 1 draw lines parallel to the 
asymptotes meeting in I. ; repeat this for the points 2, 2, 
obtaining II., and for the points 3, 3, obtaining III., and 
so on. 

A fair curve through the points I. II. III. ... is the 
hyperbola required. 

This construction is based on Theorem 3, Art. 105. 

Example. Copy Fig. 109 half as large again as shown. 



108 PRACTICAL PLANE GEOMETRY chap. 

110. Miscellaneous Examples. 

*1. CD, CB are conjugate semi-axes of an ellipse. /'J/, is drawn 
perpendicular to CD, and PN (on PM or MP produced) is 
made equal to CD. Show by construction that if the line MN 
slides between the lines CN and CD, the point P on il will 
trace out the ellipse. ('889) 

Hint. Mark the points N, P, M on tracing-paper ; as this 
template moves in the given manner, prick through at /'. 
*2. O is the centre of an ellipse, whose major axis lies on AB and 
is 3.30" in length. CD is a tangent to the ellipse. Draw the 
half-curve above AB. ^895) 

Hint. Find the foci by applying Theorem 2, Art. 91. 
*3. OH, OK, are tangents to an ellipse at G and E respectively. F 
is one of the foci. Find the axes and draw the curve. (1897) 
Hint. See Ex. 6, p. 95. 
*4. Determine the axes of the ellipse which would touch ab and have 
F and F as foci. Without drawing the ellipse find the points 
where the line through /' perpendicular to ab would cut it. 
*5. Construct the ellipse which has the given points/ 1, and F' as foci, 
and which touches the given line ab. ( T 89o) 

Hint. Use Theorem 2, Art. 91. 
*6. -S" is one focus of an ellipse. Q and B are two points on the 
ellipse, and PT is the direction of the major axis. Find the 
other focus (geometrically) and draw the ellipse. (1896) 

Hint. On BQ prbduced find D such that DQ : DB 
= SQ : SB. Draw DZ perpendicular to ST, then DZ is the 
directrix. The eccentricity may now be found, and therefore 
any number of points on the ellipse. Thus the vertices may 
be found. 
*7. /"is the focus of a parabola, AB is a tangent to the curve, and 
A is on the directrix. Find the axis and directrix, and draw a 
sufficient portion of the curve to show that AB is truly tangent 
to it. (1893) 

Hint. Apply Theorems 2 and 10, Art. 100. 
8. Draw two lines, AB, AC, including an angle of 30 . On AC 
set off a point P, so that AP=t,". AB is the axis of a para- 
bola ; AC is a tangent to the curve at the point /'. Find the 
focus and directrix and draw the curve. ( J 894) 

Hint. Use Theorems 3 and 5, Art. 100. 
*9. The directrix, and two points C and D on a. hyperbola, are 
given. If the eccentricity be J, draw the branch of the curve 
on which the points lie. (1898) 

Hint. A focus is readily found. 



IV 



CONIC SECTIONS 



109 



B 



Cb/iu She figures double sue 
A a 



9 



4 and 5 



o 

C 



'D 







D 




E 



A 



F 
7 



F' 



K 



o 

F 



F 



R 



Q 





CHAPTER V 



SPECIAL CURVES 



111. Roulettes. Definitions and construction. The 

last chapter dealt with the very important curves known as 
the conic sections. We have now to consider some other 
well-known curves, confining attention to those which have 
some application to the work of the engineer and architect. 
Taking the curves in the order of their importance from 
this point of view, the first perhaps to call for discussion 
are those which are generated by the rolling of one curve 
on another. The teeth of wheels generally have profiles 
thus formed. 

.Definitions. If two curves roll upon one another without 
sliding, any point connected with one traces upon the 
plane of the other a curve which is called a roulette. One 
curve is generally fixed or regarded as fixed, and is known 
as the base or fixed curve or directing curve ; the other, 
which then rolls over the base, is called the generating 
curve or the rolling curve. 

General method of construction} Let A A be the rolling 
or generating curve carrying the tracing-point P, and let 
BB be the fixed curve or base. 

By the use of a French curve, or otherwise, draw the 
curve BB in ink on the paper ; draw the other curve A A 

1 See paper by Mr. W. I. Last, on the " Setting out of Wheel Teeth," 
Mm. Proc. Inst. C.E. vol. Ixxxix. 1887, p. 335. 



chap, v SPECIAL CURVES in 




in ink on stout tracing-paper, or with a needle-point on 
thin transparent celluloid, and mark any point P on the 
tracing-paper or transparent template. Now adjust the 
template until the curves touch each other, and prick off 
the point P. Next insert the pricker at the point N 
where the curves appear to touch, and rotate the template 
through a small angle into a new position, and again prick 
through at P. Then insert the pricker at the new point 
of contact N, rotate slightly, and prick off P. The locus 
of P is the roulette PL, and the above process is con- 
tinued until any desired portion of this curve is obtained. 

The procedure may be varied by drawing BB on the 
transparent template, and AA and P on the paper. Then, 
operating as before, the roulette will appear on the moving 
transparency. This example affords a good illustration of 
relative motion. 

The method just described does not ensure absolutely 
pure roiling of the curves A A and BB on one another ; 
but by taking the steps sufficiently small, the errors come 
well within the limits specified in Art. 2, and are not 
measurable. Practically, the result is found to be very 
perfect, and the roulette is obtained with a quickness and 
precision far exceeding that given by any ordinary geo- 
metrical construction. 

Note. If the template be long and of tracing-paper, a strip or 
drawing-paper glued to it will prevent change of shape by buckling. 



H2 PRACTICAL PLANE GEOMETRY criAP. 

112. Problem. To draw the normal at any point P 
of a roulette, and to find the centre of curvature. 

The normal and centre of curvature have been denned 
in Art. 90 in reference to any curve. 

In the present case let the transparency be placed 
with the tracing-point coinciding with the given point P 
on the roulette, and let this coincidence be maintained 
with the pricker inserted at P, while the template is 
turned until the curves A A, BB come into contact at N. 
Then PN is the normal at P. 

For at the instant that the roulette at P is being described the 
pricker is at N, and the template is then turning about this point. N 
is called the instantaneous centre Of rotation. The tracing- 
point at this instant is thus moving at right angles to PN. 

The centre of curvature must be on the normal PN. 
To find its position, let CD be the common normal to the 
curves in contact at N. Let C be the centre of curvature 
for the point N on AA, and D that for N on BB. Join 
PC, and through N draw a line perpendicular to PN. 
Let these lines intersect at Q. Join QD to intersect the 
normal PN m O. Then O is the centre of curvature for 
the point P on the roulette PL. 

113. Problem. (a) To find the length of the given 
circular arc AB. (b) To set off a circular arc AB equal to 
the given line AE. (c) To mark off a circular arc AD 
equal to the given circular arc AB. 

The line AE must towh both the arcs at A. 

(a) Divide the arc AB into four equal parts at r, 2, 3. 
Set off AR on the tangent equal to the chord A\. With 
centre R, radius RB, describe the arc BE. Then AE is 
equal in length to the arc AB nearly. 

(b) Make AR equal to \ AE. With centre R, radius 
RE, describe the arc EB. Then AB is the arc required. 

(c) Find R as in (a). With centre R, radius RB, 
draw arc BD. Then arc AD = arc AB = AE nearly. 

Note. These constructions are only approximate. They should not 
be used for arcs which subtend angles greater than 90 . 



SPECIAL CURVES 



"3 




Examples. 1. Draw a. fine line in ink on drawing-paper, and a 
circle about 2" diameter with a fine ink line on stout tracing- 
paper. Roll the circle exactly once round on the line, 
operating with the pricker. Measure carefully the lengths of 
the circumference thus obtained, and the diameter of the circle, 
and calculate the ratio of the two. Ans. 3.14. 

2. Repeat this, but draw the line on the tracing-paper, and the 

circle on tbi drawing-paper. Ans. 3.14. 

3. Draw a circle 3" diameter. Find the length of \ of its circum- 

ference by Prob. 1 13. Measure this length and that of the 
diameter, and calculate the ratio of the two. Ans. . 7^5- 

4. Draw a circle of 3" radius. By Prob. 113 set off on it an arc 

equal to the radius, and measure the angle subtended by the 
arc at the centre of the circle. Ans. 57.3. 



114 PRACTICAL PLANE GEOMETRY chap. 

114. Cycloidal Curves. When the two curves which roll 
on one another are circles, the roulettes form a class known 
as cycloidal curves. 

If the tracing-point be on the circumference, we have : 
a cycloid when the circle rolls on a straight line ; an epi- 
cycloid when it rolls on another circle externally ; and an 
hypocycloid when it rolls internally. 

If the tracing -point be not on the circumference, 
then, when the circle rolls on a straight line, we obtain a 
prolate cycloid when the point is inside, and a curtate cycloid 
when the point is outside the rolling circle. These two 
varieties are also called trochoids. Thus epi- or hypo-trochoids 
result when the circle rolls outside or inside a fixed circle. 

We shall now give some of the usual geometrical con- 
structions for setting out these curves. It should be 
understood, however, that they are distinctly inferior, as 
regards expedition and accuracy, to the general method 
explained in Art. in, in which a transparent template 
is used. 

115. Problem. To describe a cycloid, having given the 
rolling circle. 

Let the circle to the left of the figure be the rolling 
circle in its initial position, the tracing-point P being then 
at o. 

Divide its circumference into a number of equal parts, 
say twelve, and draw a tangent to the circle at its lowest 
point. By Prob. 113 make 02' equal to the circular arc 
02, and step off 02' six times from o to o'. Bisect these 
lengths, thus obtaining twelve equal divisions, each equal 
to one of the twelve equal arcs round the circle. 

As the circle rolls to the right, the points of division 
on its circumference occupy, in turn, the positions indi- 
cated by the corresponding points of division on the 
tangent 00'. Also the tracing-point P wi'l ascend and 
will cross, in turn, the horizontal dotted lines drawn 
through 1, 2, 3, . . . 6, afterwards descending. 



SPECIAL CURVES 



ii5 




Ok' 



For example, when rolling takes place, the point 2 
descends to 2', and at the same time the tracing-point J } 
ascends to the level indicated by the dotted line through 2. 

But 2'P. Z - chord 02, hence the construction is as 
follows : 

Take the points o, 1', 2', . . . 6', in turn, as centres, 
and describe arcs to intersect the horizontal lines through 
the corresponding points on the circle, the radii being the 
lengths of the chords from o to the corresponding points 
on the circle. 

The second half of the curve may be readily obtained 
from the first half from considerations of symmetry. The 
two halves are symmetrical with respect to the line -P G 6'. 

116. Examples on Problems 114 to 119. 

1. Construct a cycloid, the diameter of the rolling circle being 2.5". 
Select any point P on the curve and draw the normal at P and 
find the centre of curvature. 

2- Construct the epicycloid and the hypocycloid, the diameters of 

the fixed and rolling circles being 4^" and 1-^" respectively. 
Select a point on each curve, for which determine the normal 
and centre of curvature. 

3- Two circles of 4" and 2" diameters have internal contact. By 

the method of Art. in set out the curves traced by a point in 
the circumference of one on the plane of the other. Also con- 
firm the fact that any point in the plane of the 2" circle will 
trace an ellipse on the plane of the 4" one. 



n6 PRACTICAL PLANE GEOMETRY chap. 



117. Problem. To construct an epicycloid, having 
given the radii of the rolling and fixed circles. 

Let C x be the centre of the fixed circle, and C that of 
the rolling circle in its initial position, the point of contact 
o being the initial position of the tracing-point P. 

The construction given for the cycloid, Prob. 1 1 5, 
applies to this problem with the exception that in place of 
the straight lines drawn through the points of division on 
the circumference of the rolling circle, there will be a 
series of circular arcs concentric with the fixed circle. 

118. Problem. To construct a hypocycloid, having 
given the radii of the rolling and fixed circles (Fig. 117). 

Let C x be the centre of the fixed circle, and C. 2 that of 
the rolling circle in its initial position. 

The construction, being identical with that of the last 
problem, except that the rolling circle is now inside the 
fixed circle, is not shown, but the curve is drawn. 

119. Problem. To determine the tangent, normal, 
and centre of curvature for any point P on a cycloid^ 
epicycloid, or hypocycloid (Figs. 115 and 117). 

Art. 112 should be read again. 

First determine the position of the rolling circle when 
the tracing - point is at P. To do this, with centre P, 
radius C n 6, describe an arc to intersect the locus of the 
centre of the rolling circle in C. With C as centre, draw 
the rolling circle through P, and find N, its point of con- 
tact with the fixed line or circle. Join PN, and draw 
PT perpendicular to PN. Then PN is the normal and 
PT the tangent at P. 

To determine O, the centre of curvature, draw from P 
the diameter PCQ. 

For the cycloid draw through Q a line perpendicular 
to the fixed line to intersect the normal PN produced in O. 

For the other curves join Q to the fixed centre C x to 
itnersect the normal in O. 



SPECIAL CURVES 



117 




120. Special cases of cycloidal curves. There are two 
special cases with which the student should be acquainted. 

In the first (a) the hypocycloid is a straight line, being a 
diameter of the fixed circle. This occurs when the diameter 
of the rolling circle is half that of the fixed circle. 

In the second case (b) the circles have internal contact, 
but the rolling circle is larger than the fixed circle. The 
resulting curve is an epicycloid, identical with that which 
would have been described by a circle of diameter equal 
to the difference of the two diameters, rolling outside the 
fixed circle, as shown at C . 



n8 PRACTICAL PLANE GEOMETRY chap. 

121. Peculiarities exhibited by curves. We shall now 
illustrate some special features which the student may find 
in the curves which he has occasion to draw, and give 
the names by which such are known to mathematicians. 

(a) Asymptotes. A straight line is said to be asymptotic 
to a curve when, if the two recede to an infinite distance, 
they get nearer and nearer together without limit, but never 
actually coincide. Similarly two curves are asymptotic 
when they approach nearer and nearer together without 
limit, but never actually coincide, as the lengths of the 
curves increase without limit. Thus a spiral may be asymp- 
totic to a circle, see the figure. 

(b) Nodes. If two branches of a curve cross one 
another, the point where they cross is called a node. Thus 
in the figure the two branches cross at the node and unite 
to form the loop. There are two tangents to the curve at 
the node, one to each branch. A node is otherwise known 
as a double point. If three branches intersect at a point, we 
have a triple point, and so on. See the figure. 

(c) Crisps. If a point when tracing a curve come to a 
place where it stops for an instant and then returns on 
itself, so as to generate two branches which have a 
common tangent at the point, we have a cusp. Two 
varieties of cusps are shown in the figure. 

(d) Points of inflexion. If a curve cross its tangent at 
any point, the latter is called a point of inflexion. Several 
other alternative and equivalent definitions might be given. 
Thus a point of inflexion is where three consecutive points 
on the curve are in a straight line, or where the radius of 
curvature is infinite, the centre of curvature crossing over 
from one side of the curve to the other, or where the curve 
changes from the concave to the convex on the same side, 
or vice versa. Or where the tangent accompanying a point 
which traces the curve has its direction of rotation reversed, 
being stationary for an instant at the point of inflexion. 



SPECIAL CURVES 



119 







(C) 




(d) 



120 PRACTICAL PLANE GEOMETRY chap. 

122. Envelopes. If a curve move in any definite manner^ 
there is a certain curve which it always touches ; this is 
called the envelope of the moving curve. 

Thus let successive adjacent positions of the moving 
curve be those marked i, 2, 3, 4 . . . Then a fair curve 
drawn to touch these is the envelope of the moving curve. 

Let the curves I and 2 intersect in a; 2 and 3 in b ; 3 and 4 in c, 
and so on. Then the envelope may also be defined as the curve 
through the intersections a, l>, c . . . of consecutive curves, when 
these are taken very very near to each other, or when the curve is 
moved by indefinitely small steps. 

The reader may acquire a very clear notion as to the nature of an 
envelope by proceeding thus : Take a piece of transparent sheet 
celluloid, and shape one edge of it to the form of the moving curve. 
This is most readily effected by tracing the curve on the celluloid by 
means of a French curve and needle-point, then bending and breaking 
the celluloid along the scratched line. Now let the manner in which 
the curved edge of the celluloid shall move be decided in the following 
way : On the celluloid draw some curve, say a circular arc, and on 
the drawing-paper draw another curve, say another circular arc. Make 
the first arc roll on the second as explained in Art. m, but instead of 
pricking through at /"use the curved edge of the celluloid as a template, 
and by means of it draw a curve on the drawing-paper for each suc- 
cessive position. 

One method of setting out wheel teeth is by envelopes. 

Example. Find the envelope of a straight line which moves so 
as to have two points P and Q in the line, 3" apart, always on 
two lines pp and qq, which cross one another at 6o. 

123. Parallel curves. If a series of normals of equal 
lengths be drawn from consecutive points on any curve, 
the curve through their ends is said to be parallel to the 
first curve, or the two curves are parallel to one another ; 
they are equidistant at all points when measured normally. 
Otherwise, if a circle move with its centre on any curve, 
the envelope of the circle is a curve parallel to the first 
curve. This second definition indicates the practical 
method usually adopted when drawing parallel curves ; in 
applying it, the student will see that the circle has two 
envelopes, thus giving two parallel curves on each side of 
the original curve. 



SPECIAL CURVES 



121 



% 5 Ar b 6 




A curve parallel to a circle is a concentric circle, and 
that parallel to a straight line is a second straight line. In 
all other cases the parallel curve is of a different character -from 
the original one. Thus the parallel curve to an ellipse is 
not an ellipse. This is seen very clearly in the figure. 
Here the middle curve, an ellipse, is first drawn. Then a 
series of circles are drawn of constant radius, with their 
centres on the ellipse, and sufficiently near together to 
enable the envelope to be drawn freehand. The external 
envelope is not an ellipse, though it might appear so to an 
unpractised eye. But no one could mistake the inner 
envelope for an ellipse, with its four cusps and two nodes. 

Note. A cusp occurs on the envelope for a point on the ellipse 
where the radius of curvature is equal to the radius of the moving 
circle. The outer envelope never has cusps ; the inner envelope only 
has cusps when the radius of the circle lies between the greatest and 
least radii of curvature of the ellipse which occur at the ends of the 
minor and major axes. If A', r be their lengths, a and b the semi- 
axes, then r, b, a, A are in geometrical progression. Hence given a 
and b, we can find A and r. 



122 PRACTICAL PLANE GEOMETRY chap. 



124. The involute and evolute of a plane curve. 
Definition \. If a straight line roll on a curve, the locus 

of any point on the line is called an involute of the curve. 

The involute is thus a special case of a roulette, and can be set out 
in the manner explained in Art. III. 

Definition 2. The locus of the centre of cmvature of any 
curve is called the evolute of the curve. 

Thus let P v /% /g ... be consecutive points on a curve ; P l O v 
P. 2 2 , P 3 O s . . .' the normals ; and O v O.,, 3 ... the several 
centres of curvature ; then the fair curve through O v 0%, <9 3 . . . is 
the evolute of the curve P v P 2 , P 3 . . . 

It will readily be seen that corresponding to a given curve there is 
only one evolute, but an infinite number of involutes. 

It is shown in works on pure mathematics that any 
normal PO to the curve at T is a tangent to the evolute at 
O. Also that the difference between any two radii of 
curvature is equal to the corresponding arc of the evolute ; 
i.e., P % Oi-P x O x = *rc 0,0,; or P 3 3 - P x O x = arc 0,O z , 
and so on. Thus we have the following theorems : 

Theorem 1. The evolute of a curve is the envelope of the 
normals to the curve. 

Theorem 2. If the evolute 00 of a curve PP be drawn, 
then PP 'is one of the involutes of 00, and might be traced 
by a point P in a straight line which rolls on 00. 

Theorem 3. Pi a roulette which is traced by a point P in a 
straight line which rolls on any curve, the point of contact O is 
not only the instantaneous centre of rotation, but also the centre 
of curvature for P. 

125. Problem. To draw an involute of a given circle, 
and to find the tangent, normal, and centre of curvature 
for any point P on the curve. 

Divide the circumference into a number of equal parts, 
say eight. At the division o draw a tangent, and by the 
aid of Prob. 1 13 set off 02' equal to one-quarter the circum- 
ference. Step off 02' along the tangent, and subdivide 
each of the steps as shown, thus obtaining divisions on the 
tangent equal to the arcs of the circle. 



SPECIAL CURVES 



12- 




Draw tangents to the circle at the points, and along 
each tangent set off the distance of the corresponding point 
on o8' from o. For example, make 44 1 = 04'. 

The curve may extend indefinitely outwards. 

Let P be any point on the involute. From P draw a 
tangent PN to the circle, and find N its point of contact. 

Then PN is the normal at P, N is the centre of 
curvature, and PT, perpendicular to PN, is the tangent. 

This construction is inferior to the method described in Art. 



1 1 1. 



124 PRACTICAL PLANE GEOMETRY chap. 

126. Spiral curves. Let a straight line, starting from a 
position OX, rotate continuously in one direction about 
O, and at the same time let a tracing- point P move 
continuously along the line always receding from or 
approaching towards O ; then the curve generated by P is 
called a spiral. The point O is called the pole ; OX is the 
initial line; OP the radius vector (or radius), and the 
angle XOP is called the vectorial angle, for the point P on 
the curve. 

If during the tracing of the curve the revolving line 
make one, two, or more rotations, the spiral is said to 
consist of one, two, or more convolutions. 

The nature of the curve depends on the law connecting 
the motions of P along the line and the line itself round 
O ; there is evidently an infinite variety of form. 

We shall give the construction for two of the best known 
spirals, the laws for which are simple. 

In the first, equal amounts of increase in the vectorial 
angle and radius vector accompany one another, or, if the 
vectorial angles are in arithmetical progression, so are also 
the radius vectors. This is the Spiral of Archimedes. 

In the second, if the vectorial angles increase by equal 
amounts, that is, form a series in arithmetical progression, 
the radius vectors form a series in geometrical progression, or, 
the ratio of any two radius vectors differing by the same 
angle is constant. This is the Logarithmic Spiral. 

The involute of a circle is a spiral curve. 

127. Problem. To draw an Archimedian spiral of two 
convolutions, having given the pole 0, and the longest and 
shortest radii OA, OB. 

Take OABX as the initial line. 

From O draw a series of lines at equal angles with one 
another, say 30 , for which angle the lines can all be 
drawn with the 6o and 30 set-square, and there will be 
twelve radiating lines altogether. 

Bisect AB in C, and divide AC and CB each into 



SPECIAL CURVES 



125 




2 78 24- X 



twelve equal parts. Then AC or CB is equal to the 
increase in the radius vector per revolution, and each of the 
divisions is equal to the increase for 30. 

Therefore make 01. = Oi, Oil. = O2, Olll. = 0$, 
and so on. Then a fair curve through A, I. II. III. . . . 
B is the Archimedian spiral required. 



Examples on Problems 126 to 129. 

1. Draw two convolutions of an Archimedian spiral, the radius 

vector increasing from .5" to 2.5" in the two turns. 

2. Draw a logarithmic spiral of two convolutions, the least radius 

being j", and the ratio of two radii at an angular interval of 
22^ being 1.08. 

3. Set out a logarithmic curve, taking 16 equidistant ordinates f" 

apart ; the least ordinate being J", and the ratio of any two 
consecutive ordinates 9:10. 



126 PRACTICAL PLANE GEOMETRY chap. 

128. Problem. To construct a logarithmic spiral 
having given the ratio of the lengths of any two radii 
and the angle between them. Let the ratio be 9 : 10 
and the included angle 30. 

From any pole O draw a series of radials making 30 
with one another. This is best done with the 30 and 
6o set-square and tee-square. 

On any line (preferably one drawn with the tee-square) 
mark off oa /line units long on any convenient scale. At 
a erect a perpendicular to oa. With centre 0, radius ten 
units (on the same scale), draw an arc intersecting the 
perpendicular in b. Join ob and produce this line. 

Now draw in succession : be perpendicular to ob ; ed 
perpendicular to oc . . . ; and ax perpendicular to ox ; 
xy perpendicular to oy . . . We thus obtain a series of 
lines . . . oz, oy, ox, oa, ob, oc, od, . . . whose lengths are 
in geometrical progression, the ratio of any two consecutive 
terms being 9:10. 

Let these lengths be set off in succession along the 
radii from the pole O, i.e. make OZ= oz, OY oy, OX=ox, 
OA = oa, OB = ob, and so on. 

The fair curve through . . . ZYXABC ... is the 
logarithmic spiral required. 

One property of this spiral is that the tangent P'P at any point P 
makes a constant angle with the radius vector OP wherever P may be ; 
the curve is therefore also known as the equiangular spiral. 

129. Problem. To draw a logarithmic curve. 

Draw any straight line OX, and along it, by applying 
the scale, or otherwise, mark off a series of points, 
1, 2, 3, 4, . . . at equal distances apart, and erect per- 
pendiculars at the points. Set off on the perpendiculars 
in succession a series of lengths which are in geometrical 
progression, found as in the last problem. 

The fair curve through the ends of these perpendiculars 
is the logarithmic curve required. 

The base XO is an asymptote to the curve. 



SPECIAL CURVES 



127 




/ Z 3 4- 5 6 7 8 9 10 77 7Z 



128 PRACTICAL PLANE GEOMETRY chap. 

130. Problem. To draw a curve of sines of given 
amplitude. 

With centre O, and radius equal to the given amplitude, 
describe a circle. Draw OF and take any two points D, 
E on this line. 

Divide the circle and DE into any and the same 
number of equal parts, say 16, numbering the parts in 
each case from o to 16. 

At any point, say 3, on DE erect a perpendicular to 
meet a line drawn parallel to OF from 3 on the circle ; 
this gives one point on the required curve ; repeat this 
construction for each of the remaining points, and finally 
draw the fair curve through the points on the perpen- 
diculars. This is the curve required. 

A point of inflexion occurs where the curve cuts DF. 

This curve is the same as the projection of a helrx or 
screw thread. See Prob. 369. 

The ordinate 33 of the curve is proportional to the 
sine of the angle AO3, since the sine is equal to the 
ordinate t>2> divided by the radius of the circle. See Art. 
13. Hence the name of sine curve or curve of sines. 

131. Simple harmonic motion. Definitions. 

Suppose a point P to move at a constant speed in a 
circular path ; and let PM be a perpendicular from P 
to a fixed diameter AB. If this perpendicular move with 
P, its foot M is said to execute a simple harmonic motion or 
simp/e vibration. The circle in which Amoves is called the 
directing circle, and P v~> the directing point. 

The amplitude of the vibration is equal to CA or CP; 
that is, to half the travel of the vibrating point M, or to 
the radius of the directing circle. 

The period of the vibration is the time which elapses 
while P goes once round the directing circle ; or it is the 
time occupied in a complete vibration of M, there and 
back. 

The phase of the vibration for any position M is the 



SPECIAL CURVES 



129 



E/ F 




fraction of the period which has elapsed since the moving 
point last passed through its middle position in the positive 
direction. Thus in the figure let from A to B be the 
positive direction, and let the rotation of CP be clockwise, 
as indicated. Then the phase for the position of J/shown in 
the figure is the angle DCP, expressed as a fraction of one 
revolution. Thus to obtain the phase we might measure 
the angle DCP in degrees and divide by 360 . For a 
phase of |th, DCP is 45. If M were at A, the phase at 
that instant would be f . 

Examples. 1. Set out a curve of sines the amplitude OA being 

ii" and the base DE 6" long. 
2. Measure the series of equidistant ordinates or perpendiculars for 

that portion of the curve situated above the base DE, Ex. I. 

Calculate the mean ordinate by Simpson's second rule, Prob. 

43. Ans. .95". 

K 



130 PRACTICAL PLANE GEOMETRY chap. 

132. Component and resultant motions. 

We now consider the path of a point which has two or 
more motions simultaneously given to it. The motion which 
the point actually has is called the resultant, while the inde- 
pendent motions giving rise to this are named components. 

In order to fix our ideas, let us suppose that a point 
moves uniformly towards O over a length PQ of the bar 
AO, while the bar rotates uniformly about a fixed point O 
through the angle AOA' ; then the point will arrive at Q 
by way of a certain curve PBQ. The point may, however, 
be supposed to move from the position P to that of Q' in 
either of the two following ways amongst others : 

First, let the point move from P to Q while the bar 
remains in the position AO ; then allow the bar to rotate 
through the angle AOA' ; thus the point arrives at Q. 

Secondly, let the bar move from AO through the angle 
AOA' without allowing the point to move along the bar, 
P will move to P' ; then let the point move on the bar 
from P' to Q'. The point again reaches Q'. 

In each of these two latter ways the point receives its 
component motions in succession. Although the paths are 
different, the final position Q' is the same in all three ways. 

Now let us further suppose that, in addition to the two component 
motions already referred to, the point receives a third component 
motion, due to O having a uniform motion from O to O' along 00', 
the three taking place simultaneously ; in this manner the point will 
arrive at Q" by way of a certain curve PB'Q". 

The point may, however, be brought from the position r to that 
of Q" by allowing it to receive the first two component displacements 
in either of the above two ways, and then allowing O to move along 
00' to O', the bar moving from OA' to O'A" without rotation ; thus 
the point will arrive at Q". By giving the component displacements 
in succession in every possible order it is seen that there are six ways 
in which the point may arrive at Q". From these observations the 
truth of the following statement will be manifest : 

If a point has a number of simultaneous component motions 
impressed upon it, it can be brought from any o?ie of its positions 
to any other, by giving to it its corresponding component displace- 
ments in succession, and in any order. 



SPECIAL CURVES 



13 1 



Q 





\ ^""^ia. / 

\ ?P? 

\B / F 

\ / 
/ 

/ 
1 / 


A' 














' 




^C 




/ 
/ 
/ 
/ 
/ 
/ 
/ 

132 


/ 






/ 
/ 
/ 
/ 
/ 



tf' 




^ /' 2' J' <' J' ' 7' ^> 



133. Problem. A point P moves backwards and for- 
wards at a constant speed between two points A and B 
in a straight line, and the line has a similar motion 
between given limiting positions ab and a'b'. The time 
of the first oscillation is double that of the second, and 
the point starts from the position a. Determine the path 
traced by the point. 

Divide ad into a number of equal parts, say four, and 
divide ab into double the number of equal parts, that is 
eight Draw the dotted lines as shown. 

From the conditions it is evident that when AB occupies 
the. position n say, the point /"will be in the line iV, so 
that P x will be the actual position of P at this instant. 

Proceeding in this way, the zigzag path a^'bxa is 
obtained, as required. 



i 3 2 PRACTICAL PLANE GEOMETRY chap. 



131 Problem. A point M has two component simple 
harmonic motions of the same period in directions at right 
angles to one another ; to trace the curve described by the 
point, having given the amplitudes, and the phases at any- 
instant. 

Refer to Art. 131 for definitions. 

Draw any two perpendicular lines intersecting in T, in 
which take points Cand C. It is convenient to make TC 
and TC each equal to the sum of the given amplitudes. 

With centres C and C describe circles with radii equal 
to the given amplitudes, and draw the diameters AB, A'B' 
perpendicular to TC, TC as shown. 

Choosing as positive directions those indicated by the 
arrows on the diagram, viz. the directions from A to B and 
A' to B', with clockwise rotations, then D and D' are the 
positions of the directing points which correspond with zero 
phase. Now set off the angles DCB, D'C'P', clockwise, 
fractions of one revolution corresponding to the given 
phases. In the figure, the angle D'C'P' is for a phase of 
T Lth, and is made equal to T V of 360 or 36 . The 
angle DCP is 105 , the phase being ifg- or J T . 

Next divide the two circles into the same number of 
equal parts, say 1 2, P and P being the zero points, and the 
points of division o, 1, 2 . . ., o', 1', 2' . . ., proceeding clock- 
wise. From corresponding points draw lines respectively 
parallel to CT and C'T; their intersections will give points 
on the required curve. A pair of these, those from the 
points 3, 3', are shown intersecting in Af.,, the others being 
omitted to avoid confusing the figure. 

The curve is an ellipse, and the circumscribing rectangle 
should be drawn as shown, the points where the ellipse 
touches the rectangle being determined. Thus to find the 
point of contact M. When the tracing-point is at M, P is 
at A, between 5 and 6, and P is at m between 5' and 6'. 
Therefore find m by setting off the angle 5' Cm equal to the 
angle 5CA, then from m draw a line parallel to C'T to 
intersect the side of the circumscribing rectangle in M. 



SPECIAL CURVES 



133 




Examples. 1. A point M has two component simple harmonic 
motions of the same period and equal amplitudes of \V' , in 
directions at right angles to one another ; set out to scale the 
curves traced by M, the initial phases being (a) o and o ; (//) 
o and \ ; (c) o and y^. 

Atis. (a) a straight line ; (/') a circle ; (c) an ellipse. 

2. A point M has two component simple harmonic motions in 

perpendicular directions. The amplitudes are i" and i^" ; 

periods I to 2 ; and initia phases (a) o and O ; (fi) o and j ; 

(() o and 1. Set out the paths of M. 

Ans. (a) Double-looped curve having a node and a point 
of inflection on each branch at the centre; (/>) a para- 
bola. 

3. Determine the curves traced by a point I\I which has two 

component simple harmonic motions at right angles as follows : 
Amplitudes l" and i\" ; periods 2 to 3 ; initial phases 
(a) o and O ; {/>) o and ^ ; (<) o and T V 

4. Trace the curves described by a point M which has two com- 

ponent simple harmonic motions in perpendicular directions of 
amplitudes 1" and 1^" ; periods 3 to 4 ; and initial phases (a) 
o and o ; (l>) j and \ ; (c) j and o. 



134 PRACTICAL PLANE GEOMETRY chap. 

135. Problem. A point starting from the position P 
moves uniformly round the circumference of the circle, 
centre C, while the circle turns about the fixed point in 
such a manner that the diameter OA moves through the 
angle AOA' of 90 and back again without stopping with 
uniform angular velocity ; determine the locus of the point 
P. 

With centre O draw the arc AA' . 

Divide the circumference of the circle, centre C, into a 
number of equal parts, say twelve, and the arc AA' into 
half the number, six. 

The method adopted is that of finding a series of 
positions of P, giving a succession of points on the curve, 
and we shall illustrate the method by determining the 
position of the tracing-point when it has moved over one- 
twelfth of the circumference of the circle. 

Let the point receive its component motions in succes- 
sion, Art. 132. First suppose that it receives its circular 
motion about C ; this will carry it from jP to 1. 

Next, suppose that the circle turns about O until OA 
arrives at 0\ ; during this second motion the point turns 
about O, moving from 1 to P v where the angle iOJ\ = 
angle AOi. 

P Y may be determined from the fact that the triangle 
Oi'P 1 is equal in all respects to the triangle OAi. 

Proceeding in this way, the series of points may be found, 
and a fair curve drawn through them. 

The path from P 6 to P is a straight line. 

136. Problem. OX is a vertical axis, and OA the 
initial position of a rod which turns about in a plane 
containing OX until it has described an angle which is 
double the angle AOX, the point during the same time 
moving along OX to 0'. If both of these motions be uniform, 
determine the locus of a point which starts from and 
moves at a constant speed along the rod from to A and 
back again in the time that moves to 0'. 



SPECIAL CURVES 



135 




136 PRACTICAL PLANE GEOMETRY chai\ 

The point has three component motions : (1) a motion 
along the rod ; (2) an angular motion about O with the 
rod ; (3) a motion due to the point O sliding down 00'. 

Divide 00' into a number of equal parts, say eight ; 
also divide OA into half the number, that is four equal 
parts. With O as centre describe the arc A4', and divide 
this arc into four equal parts. In each case number the 
points of division as shown. 

We shall show how to determine the position of the 
tracing-point when one-fourth of the total time has elapsed ; 
that is, when O has moved to the point 2. 

Taking the component displacements in succession : 

1. Suppose the bar to have its vertical motion bringing 
it to the position 02, where Aa =02. 

2. Let the bar then have its angular motion ; that is, 
with 2 as centre describe the arc aA. making liA. 1 = A2'. 

3. Now let the point move along the bar ; that is, make 
2P 2 = O2 . Then P 2 is the desired point. 

Repeat this construction for the eight positions. 

The component motions might have been given in any 
order, but probably that indicated will involve least trouble. 

The path of P is shown. 

137. Motions under mechanical constraint. All the 
curves and in fact the lines of all the geometrical figures 
we have yet considered may be regarded as having 
been described by points moving under some kind of 
constraint. A straight line, for example, by a pencil point 
guided by a straight-edge. A circle by a point controlled 
by a second point (or axis) embedded in the material of the 
paper and drawing-board. In other cases we have had 
geometrical conditions imposed on the motion of the 
tracing-point, and by means of the constraints of ruler and 
compasses we have found a series of isolated points in the 
required path, and have completed the curve either by 
muscular constraint, as when drawing a fair curve by free- 
hand through the points, or by the use of a template, such 
as a French curve or bent spline, to guide the pencil. 



SPECIAL CURVES 137 



We shall now give some problems relating to the motions 
of parts of mechanisms or machines. The constraint in 
such cases may be described as mechanical. The problems 
and methods of solution do not differ essentially from those 
preceding ; it is only the form of the data that is new. We 
are not required to make the mechanism nor a model of it, 
so as to allow the moving point in it to actually trace its 
own curve ; nor are we even required to make a drawing 
of, or to be acquainted with, the constructional details of 
the mechanism. 

A mechanical constraint always has its geometrical 
counterpart, and the first thing is to realise what this is 
and how it is to be represented. In the majority of 
cases the mechanical constraints consist only of sliding 
and turning pairs of elements, represented respectively by 
straight lines and circles. The moving pieces may also be 
represented by lines. 

We thus set out to scale a line or " skeleton " diagram 
of the mechanism, putting in only such lines as are 
essential to geometrically represent the moving pieces and 
the several constraints. This figure, or a portion of it, is 
then drawn for a number of positions of the mechanism as 
the latter moves through all of its possible positions, and 
the corresponding positions of the point under considera- 
tion are noted. A fair curve drawn through the series of 
points thus determined gives the required path. 

This path is always a closed curve for a machine which 
works continuously, repeating its cycle of operations, since 
in such a machine no point can go off to infinity. This, 
however, would not necessarily be the case if we were 
finding the envelope of a moving piece instead of the path 
of a point. 

It will often happen that the piece during its motion 
assumes critical ox limiting positions, the accurate determina- 
tion of which is desirable in order to draw the curve to the 
best advantage. Illustrations will appear in the particular 
problems to which we now proceed. 



138 PRACTICAL PLANE GEOMETRY chai\ 

138. Problem. To find the path of a point in the con- 
necting rod of a direct-acting steam engine. 

The figure is a skeleton diagram representing the 
mechanism in one of its positions. CB is the crank, turn- 
ing about C, and constraining the crank pin B to move in 
a circle. BA is the connecting rod, with the end B centred 
on the crank pin, and the end A compelled to keep to the 
straight line AC by means of a slide. P is the point in 
the connecting rod whose locus is required. The problem, 
stated geometrically, would read thus : - 

A given line AB of constant length moves with one end 
B always on a given circle, centre C, and the other end on a 
straight line directed through C ; find the locus of a point 
P in AB. 

We must draw the connecting rod AB, and thus locate 
P, for a number of positions in the cycle, say twelve 
positions. To do this in the best way, divide the crank 
pin circle into twelve equal parts, beginning at O (most 
quickly effected by 30 and 6o set- square). With these 
points, o, 1, 2 . . . 1 1 as centres, and radius equal to BA, 
describe arcs cutting the line in which A moves, thus ob- 
taining the twelve positions of A which correspond to the 
twelve positions of B. The connecting rod AB is shown 
in the figure for one of these pairs of positions, viz. 1, 1' ; 
and the point P 1 is marked off at the given distance from 
one end. In determining the locus of Pihe other positions 
of the connecting rod were drawn, but for clearness have 
been omitted in the diagram. 

In this and similar problems we might with advantage again make 
use of a transparent template. Thus draw the circle and the line AC 
on the paper ; then mark the points A, P, B on tracing - paper or 
celluloid, and adjusting the template in succession to a number of 
positions, with A on the line, and B on the circle, in each case prick 
off the position of P. It is now hardly worth while to divide the 
circle into equal parts. 

Examples. 1. In a direct-acting engine the crank is I foot long 
and the connecting rod 4 feet ; find the locus of the middle 
point of the rod. Scale -|th. 



SPECIAL CURVES 



139 







2_- 


j 


s? 




A __ 


p^=^= 


1 r ^^ 


c 




\ ,5 

I 6 ' 


0'/' 


^v^ 








/7 






w" 


<f 


tf 





2. The middle point of a Gooch link of 48" radius is guided along 
the centre line of a horizontal engine. The ends A, B of the 
link are 20" apart and are connected by " open " rods AP, BQ, 
60" long, to two eccentrics CP, CQ each of radius 4" and with 
angular advance of 30 . Find the motion of the valve at half 
gear, the link-block then moving in a horizontal line DD 5" 
below the centre line. 
Solution. Draw the horizontal centre line KC. Set off angles 
KCP= 120 above KC, and KCQ the same below, and make 
CP, CQ each 4". With centre C draw the circle through P 
and Q. Divide its circumference into twelve equal parts, 
numbered o to II, the divisions o and 4 being at P and Q. 
With these points as centres, radius 60", draw twelve arcs in 
ink, o to 1 1, extending about a foot above and below K. Draw 
a horizontal line DD 5" below A". 

On tracing-paper stiffened by a lath draw an arc of 48" radius, 
on which mark A, B 20" apart, and the mid point of the arc. 

Now place this template with its convex side towards C, 
and adjust it so that its lower and upper points A and B, and 
its mid point O, lie on the arcs o, 4, and the centre line respec- 
tively. Then prick through the point > Q where the curve on 
the template crosses DD. 

Again adjust the template, O being on KC as before, but A 
and B now lying on arcs 1 and 5. Prick through D x on DD. 
Repeat this for the arcs 2 and 6, determining D 2 ; for 3 and 7, 
obtaining D 3 ; and so on. 

Measure the displacements Z> , D v D 2 . . . D n of the 
valve from the average position of D. A/is. 2.71", 3.08", 2.64", 
i-55", 0.13", -1.34", -2.46", -3.01", -2.83", -1.84", 
-0.21", 1. 51". 

(From which, by Fourier's analysis, ^=3.07 sin (6+ 57) 
+ o. 14 sin (20 + 1 03). d is the crank angle measured from 6'A". ) 



i 4 o PRACTICAL PLANE GEOMETRY chap. 



139. Problem. To set out the complete path of the 
guiding point in "Watt's simple parallel motion. 

This mechanism consists of two " radius rods " or 
" levers " CA, DB, centred at C and D, having their ends 
A and B connected by a link AB. The guiding point 
P is situated in this link, and in such a position that P 
divides the link into two segments having a ratio inversely 
as the radius rods. That is, AP ': PB = DB :AC. The' 
locus of P is a two-looped curve something like a figure 
of 8, and in the neighbourhood of the node or double 
point the two branches of the curve are very flat, being 
almost straight lines. Hence P serves as a point of attach- 
ment for a piece which moves to and fro through a limited 
range, and requires to be guided in an approximate straight 
line, such as the pencil of a Richard's Indicator, or the top 
of the piston rod in a beam engine. 

To trace the locus of P. With C and D as centres, 
describe circular arcs through A and B respectively. Then 
A and B move in these paths. 

Next trace the link APB on celluloid with a needle- 
point, marking the three points A, P, B by short cross 
lines. Move this template, placing A in a new position 
A' on the arc through A. Insert the pricker at A', rotate 
the template until B comes on the other arc at B', then 
prick off the point P at P. Repeat this operation a suffi- 
cient number of times to enable the locus of Pio be drawn. 

Or operate with compasses in the usual manner. 

Limiting positions. Observe that the highest possible 
position of A is A , where DB and BA come into one 
straight lineZ>i? ^ , and where, therefore, DA = DB + BA. 
Similarly its lowest position is A v where DA^DB + BA. 
In like manner the limiting positions of B are B\ and B 2 , 
where CB 1 = CB. 2 = CA + AB. 

Thus to find "A Q and A x describe arcs with centre D, 
radius DB + BA, to intersect the path of A in A and A v 
To find B 1 and B 2 describe arcs with centre C, radius CA 
+ AB, to intersect" the path of B in B x and B. 2 . 



SPECIAL CURVES 



iji 



>Ao 




B'\B } 



Note also that A B is normal to the curve at P Q , since 
for this position A is the instantaneous centre of rotation 
of the link. This statement may be tested by trial with 
the template, and its truth will then be quite evident. 
The end A of the link, having come to the limit of its 
movement at A , is there stationary for an instant before 
retracing its path. 

Examples. 1. In a simple parallel motion the lengths of the 
oscillating rods or levers DB and CA are 3 feet and 4 feet 
respectively, and the length of the connecting link AB is 
2 feet. The mechanism is so set that when DB and CA are 
horizontal, AB is vertical. Eind the position of guiding point 
P in the link, and set out the complete locus of P. Scale 
1" to I'. 

^^=C^-^ = 7 f2 ' = - 857 ' 
Or, geometrically : Join CD to intersect AB in Q, and make 
AP=BQ. 



142 PRACTICAL PLANE GEOMETRY chap. 



4. In 



2. Work Ex. i when the centres C and D are moved 0.35' 
nearer together horizontally, so as to give the link a slight 
inclination to the vertical when the rods are horizontal, as 
in Fig. 139. 

Ans. AP= - of 2' = 0.857' as before. 

7 

Note. The best approximation to a straight line is obtained when 
the centres C and D are so adjusted that for its working range 
the link AB deviates from the vertical equally on each side. 

3. In Ex. 1 suppose the centres C and D are both on the same 
side of the link, find the guiding point P in the link for an 
approximate straight-line motion, and trace the locus of P. 

Ans. Produce BA to P, such that 

PA : PB = DB : CA, or PA : PB-PA = DB : CA - DB ; 

that is, /M=^t 2 = 5 feet. 

1 J 

Or, geometrically: For the position in which CA and DB 
are parallel, join CD and produce it to intersect pro- 
duced AB in Q, then produce BA and make AP=BQ. 
[n a drag-link coupling the shafts are 6" apart, the drag link 
12" long, and the cranks each 30" long. Find the locus of the 
middle point of the link. Scale |th. 
Or, stated geometrically : Take two fixed points C and D, 6" 
apart, and with these as centres describe circles each of 30" 
radius. A line AB of the constant length of 12" moves with 
A on the circle C, and B on the circle D. Find the locus of 
P which bisects AB. Scale -i-th. 

140. Cams. We shall conclude this chapter with an 
example of a cam. In previous examples on mechanisms 
the problem has been the direct one given the mechanism, 
to find the consequent motion. We have now the inverse 
problem, generally more difficult given the required 
motion, to design the constraining mechanism. When a 
machine designer requires a complicated motion, he gener- 
ally has recourse to a cam or a combination of cams. 
Numerous examples are met . with in textile machinery, 
printing machinery, and in many other machines. 

One form of cam is shown in the figure. It consists of 
a flat plate with an irregular contour ABC, capable of 
rotating about an axis O, and so giving a reciprocating 
motion to a piece AH, which slides in a fixed guide K. 



SPECIAL CURVES 



142 



H 



i 



1 



: 



/r 




It is obvious that as the cam rotates, the piece AH 
will receive a rectilinear motion, the nature of which will 
depend on the shape of the curved edge ABC of the cam, 
and on how the latter rotates., whether at a constant or 
varying speed. 

For example, while the cam turns through an angle 
which brings B under the end of the sliding piece, the 
latter will move upwards through a distance equal to the 
difference of the radii OB and OA. 

Hence, supposing the cam to revolve at a uniform 
speed, we could (within limits) give any kind of motion to 
the reciprocating piece by suitably shaping the edge ABC 
of the cam. The motion will be repeated with each revolu- 
tion of the cam. 

In practice the moving piece would require to have a roller pinned 
to its lower end and bearing on the cam, in order to diminish the 
friction, and to prevent undue wear. 

A spring also is generally required to keep the roller in contact with 
the cam when the piece is descending. Neither of these is shown in 
the diagram. 



144 PRACTICAL PLANE GEOMETRY chap; 

141. Problem. It is required that a reciprocating 
piece guided by a straight slide shall move at the same 
constant speed both ways, the change of motion being 
effected without interval. Set out the form of the cam 
which rotating uniformly will produce the motion, one 
revolution corresponding to one to-and-fro movement of 
the piece, and the axis of rotation being in the line of the 
slide. You are given the diameter of the roller, and the 
greatest and least distances of its centre from the axis of 
rotation. 

Let CAB be the line of the slide, taken vertical, C the 
centre of rotation of the cam, and A and B the extreme 
positions of the centre of the given roller, which is pinned 
to the lower end of the reciprocating piece. 

Divide AB into a number of equal parts, say six. 
Divide the half-revolution into the same number of equal 
angles ; i.e. draw lines through C at angular intervals of 
30 (with the 30 and 6o set-square). On these set off 
in succession the lengths Ci', C2', C3' . . ., equal to Ci, 
C2, C3 . . . Then if the roller were a mere point the 
required shape of the cam, on one side, would be the fair 
curve through A, \, 2', 3' . . ., which from Art. 127 is 
seen to be an Archimedian spiral. 

To allow for the roller we draw a curve parallel to the 
spiral, determined as the envelope of a circle of diameter 
equal to the roller, moving with its centre on the spiral. 
See Art. 123. Some of these circles are shown to the 
right of the figure. 

The right half of the cam is the same shape as the left 
half, the line of the slide being an axis of symmetry. This 
cam is known as the heart-shaped cam. 

Examples. 1. Design a cam to give the following motion to a 
sliding piece : Rise of 3" at a uniform speed during first quarter 
revolution ; rest during second quarter ; uniform fall of 3" during 
third quarter ; rest during last quarter. Diameter of roller |" ; 
least distance of its centre from axis of cam 2". 

2. It is required that a point P shall move in a straight line with a 
speed which increases uniformly from zero during a vertical rise 



V 



SPECIAL CURVES 



145 



/ 
t 


5) 






.4 
,3 
.2 




\fr 


TV 






/j2 




/ ^*c ^ ^"~"" 






* jT^ \ 






' / ^ \ 






/ / ^ N 






5/ / ^ 


6 










I '" 






ll ' ' 






' -' / 






/ 






' ' 






1 1 -^ / 






1 \s / 






t-x \ > 






n \ / 






^ \ > 






* \ 1 






* \ / 






* \ ' 






^ \ ' 






^ V 










/ / 








*s *"*- 




^''' 


"'-- 


"6'' 





of 3" ; the motion is then to be suddenly changed, and the 
point is required to fall 3" at a constant speed, the times of the 
rise and fall being equal. Design a cam which, while rotating 
uniformly, will each revolution impart this motion to P, the 
nearest approach of P to the axis of the cam being 2". 

Hint. It is shown in mechanics that when the speed increases 
uniformly, the distance from the position of rest is proportional 
to the square of the time ; that is in this case, for the rise, pro- 
portional to the angle turned through by the cam. Now in a 
parabola the abscissa AN is proportional to the square of the 
ordinate PN (Theorem 6, Art. 100). Therefore in Fig. 103 
take A AT 3" to represent the path of P. Construct the para- 
bola AQ, and from the points I., II., III. draw lines parallel 
to (Wto meet AN in P v P 2 , P y These are the positions of 
P at intervals of 45 during the rise from A to N. 
Design a cam which shall give a rise at a uniform speed during 
the entire revolution, with an instantaneous drop at the end. 

L 



I 4 6 ' PRACTICAL PLANE GEOMETRY chap. 

142. Miscellaneous Examples. 

*1. Four equal rods, ab, be, ed, da, form a frame jointed at the 
angular points. The frame is also pivoted at a. The point c 
moves on the circumference of the given circle. Draw the 
curve traced by the point x on the bar be. ( 1S87) 

*2. Points P and Q are constrained to move uniformly along the 
given lines ab and cd. While P moves from to a and back 
again, Q moves from c to d, and while P moves from to b 
and back again, Q moves from d to e. Trace the locus of a 
point on (IP produced 2" distant from Q. (1889) 

*3. and d are fixed pivots about which the bars oa, o'b can freely 
revolve ; ab is a coupling bar connecting the free ends of oa and 
ob. Draw the complete locus traced by the centre point of 
the bar ab. (1S93) 

*4. A circle A, of diameter EF, rolls on the line CD with uniform 
motion from left to right, starting from E. Another circle B, 
whose diameter is half that of A , rolls inside the circumference 
of A, also with uniform motion, but from right to left, tarting 
at E when A begins to move. Circle B is in contact with 
circle A at F at the same time that F reaches the line CD. 
Draw the curve described by the centre of circle B. (1897) 

*5. A point P revolves round the circle with centre C on the line 
OD, with an uniform motion. The point C moves from C to 
D and back from D to C, also with an uniform motion, return- 
ing to Cat the same time that /"has completed one revolution. 
(a) Draw the path of P. (b) Supposing that at the same 
time the line OD moves round the point O uniformly, making 
a complete revolution while P revolves once round the moving 
point C; draw the path of P. (1895) 

*0. Two bars, o^a, o 2 b, are pivoted at o l and o 2 respectively. At 
i they pass through a saddle which can travel along o^a at 
fths the speed at which it can move along o. 2 b. Trace the 
locus of i. What is this curve ? (iSSS) 

7. Two lines meet at an angle of 6o. Find the locus of points in 
the interior of the angle such that the sum of their distances 
from the two given lines is constant, and equal to 2|". (1S97) 



SPECIAL CURVES 



M7 




CHAPTER VI 

CO-ORDINATES PLOTTING ON SQUARED PAPER 

143. The position of a point in a plane. Co-ordinates 
of a point. Our object in this article is to show how the 
position of a point in a plane may be defined. We shall 
illustrate the case by means of a concrete example. 

Suppose a person wishes to note the position of some 
simple object such as a boy's marble on the floor of a 
room, so that he may make a scale drawing which shall 
exhibit this position. After measuring the room he will 
require to make ttvo further measurements of the position 
of the object. There is considerable choice in the latter, 
as appears from what follows. 

To fix the ideas, let the room be rectangular, 18 feet 
long and 12 feet broad, as set out to scale in the figure. 

The position in the room of the object P may be ob- 
served in the following four ways amongst many others : 

1. By measuring the distances of P from any two 

adjacent sides of the room. For example, 
PN=t.6\ PM = 4.1'. 

2. By measuring the distances of P from any two 

corners of the room. Thus 

OP=8.6', AP= 11. 2'. 

3. By measuring the distances of P from one corner 

and from one side of the room. Say 
^=11.2', PJV= 7 .6', 



chap, vi PLOTTING ON SQUARED TAPER 



149 



B 



C 



Jca/i of Feet 



s 

1 I....I ... I....I I 



10 




4. By measuring the distance of P from one corner of 
the room, and the angle which the line from P 
to the corner makes with one of the sides ; e.g. 
OP=8.6\ AOP =28.4. 

Any one of the above four pairs of measurements com- 
pletely defines the position of the point on the floor ; any 
one pair being given, the others could be determined by 
calculation or construction. The two measurements of 
any of these pairs are called the co-ordinates of the point ; 
so we may have different systems of co-ordinates. 

Rectangular co-ordinates are illustrated by Case 1. This 
system is the one most generally useful. 

Polar co-ordinates are illustrated by Case 4. This method 
of defining position is frequently employed. 

In this chapter we shall confine attention to rectangular 
co-ordinates. The above illustration does not represent 
the most general case. It requires to be amplified so as to 
include points outside the room on the floor-level. This is 
done by the convention of positive and negative co-ordin- 
ates, defined in the next article. 

Example. The floor of a room OACB is 20 feet square. A small 
object P on the floor is 14.7 feet from the side OB and 10.3 
feet from the side OA. Make a plan of the room showing 
the position of P to a scale of J" to 1' and measure 
(a) The distances of P from and A. A/is. 17.9 ft., 1 1.6 ft. 
(6) The angles AOP and OAP. Arts. 35 , 62.75 . 
(<) The distances of P from R and C. Ans. 17.6 ft., II. I ft. 



150 PRACTICAL PLANE GEOMETRY chap. 

144. Rectangular co-ordinates of a point P in a plane. 

In the plane draw any two perpendicular lines of 
reference XX and YY' intersecting in O. '1 hese lines 
are called the axes of co-ordinates, or the co-ordinate axes, 
or simply the axes. The point O is called the origin. 

Frx>m P draw perpendiculars PA/, PN to the axes. 
The lengths of these lines are the rectangular co-ordinates, 
or the co-ordinates of the point P referred to the axes. The 
horizontal distance NP or OM, measured parallel to OX, 
is called the abscissa, or x co-ordinate, and is denoted by x ; 
the vertical distance MP or ON, measured parallel to OY, is 
the ordinate, or y co-ordinate, and denoted by y. 

These co-ordinates serve to define the position of the 
point P in the plane. It is convenient to express this 
position by writing " the point (x, y)." Thus, suppose the 
abscissa PJV or x were 3 units long, and the ordinate PM 
or y were 5 units, we might speak of P as the point (3, 5), 
being careful always to write the abscissa first. 

Suppose it were required to plot the point (3, 5) on 
paper, this could be done in three ways : 

1. Along OX set off OM 3 units long ; draw a perpen- 

dicular from M, and on it mark off MP, 5 units 
upwards ; or 

2. Along OY set off ON 5 units long; from A^draw a 

horizontal line, on which set off NP to the right, 
3 units long ; or 

3. Along OX and OY set off OM and ON 3 and 5 

units long ; through M and N draw lines perpen- 
dicular to the axes intersecting in P. 

Observe that whichever method of plotting be adopted, 
in each case we measure from the axes 3 units to the right 
and 5 units upwards, in order to arrive at the position of P. 

Suppose now the point were at P 2 , the distances of 3 and 
5 units from the axes being the same, but requiring to be 
set off to the left and dozvnwards, instead of to the right and 
upwards. In writing down the co-ordinates, how could 
this case be distinguished from the last ? 



VI 



PLOTTING ON SQUARED PAPER 



'5 1 




\-P A (3,-5) 



The convention adopted is to prefix the minus sign to 
the co-ordinates of P 2 , writing them - 2 and - 3, and 
considering each as a negative quantity. We should thus 
speak of the point P as the point ( - 3, - 5). 

We might plot the point P 2 in any one of the three 
ways described for P; but in all cases, in order to arrive 
at P 2 , we must mark off 3 units horizontally to the left, and 
5 units vertically downwards. 

Thus a negative abscissa is measured or set off horizon- 
tally to the left from the vertical axis ; and a negative 
ordinate is measured or set off vertically downwards from 
the horizontal axis. 

According to these definitions, the co-ordinates of P l 
are seen to be - 3 and 5 ; and the co-ordinates of P 3 are 
3 and- 5. 

We have thus made complete provision enabling us to 
define the position of any point on either side of either 
axis, that is, of any point in the plane of the axes. 



152 PRACTICAL PLANE GEOMETRY chap. 

145. The use of squared paper. In order to facilitate 
the plotting of points, and especially of curves determined 
as a series of points, paper may be used which is covered 
with horizontal and vertical lines at equal intervals, ruled 
by machinery, and known as squared paper. The lines 
serve as horizontal and vertical scales, making ordinary 
scales unnecessary, and points are quickly located. For 
ordinary work the lines may be one-tenth of an inch apart, 
every fifth and tenth line being distinguished by a difference 
in width or colour. 

A portion of a sheet of such squared paper is here 
shown full size; the main divisions are i" apart, and 
these (or the half-inch divisions) may be read as units, tens 
of units, hundreds. . ., or tenths, hundredths . . ., exactly 
as described in Art. 5, with regard to decimal scales. 
With care the position of a point may be plotted to within 
about the one-hundredth of an inch, if the lines are accur- 
ately ruled. Sheets of this paper 18" by n" may be 
obtained at a very moderate cost, ruled in fine and faint 
blue lines, every fifth line being broader and more con- 
spicuous than the rest, and ruled alternately in blue and 
red. The red lines are thus one inch apart, as are also the 
intermediate broad blue lines. 

When using squared paper the axes of co-ordinates may 
be chosen so as to have any convenient positions on the 
sheet, according to circumstances, and the paper may be 
placed with long edges either horizontal or vertical. 

Examples. 1. Take a sheet of squared paper, and, selecting the 
origin at a corner of a 1" square somewhere near the centre of 
the sheet, plot the following points, taking 1 inch as the unit : 

(2, 5); (6, 3); (2.5. 4-6); (1.44, 355); (-3. 4); 

(-3, -i-5). 
Note. After selecting the origin, mark the divisions to the left and 
right, and up and down, as for an ordinary scale, but with the 
additional negative values. 

2. Again selecting the origin near the centre, but taking 1 inch to 
represent 10 units, plot the following points : 
(25, 36); (-42, 25); (30.4, -40.8). 



VI 



PLOTTING ON SQUARED PAPER 



153 



1 1 _ 

, 1 1 1 

1 



I I 

I I 





3. Taking 1 inch to represent 100 units, plot the following points : 

(60, 360) ; (55, 400) ; (404, 295). 

4. Plot the points (1, 2) and (7, 10) ; join them by a straight line, 

and read off the ordinates of those points on the line whose 
abscissa are 3, 6, 4.2 respectively ; also read off the abscissa: of 
those points whose ordinates are 3, 5.2, 9.6 respectively. 
Ans. 4.67, 8.67, 6.27 ; 1.75, 3.4, 6.7. 

5. Plot the points (o, 3) and (4, o) ; join them by a straight line, 

and find the length of the perpendicular drawn to the line from 
the origin. Take j" as the unit. .Ins. 2.4". 

Note. The above examples will show the student that considerable 
care and thought are necessary in selecting the position of the 
origin and choosing the scales, in order to secure the best 
results. 



154 PRACTICAL PLANE GEOMETRY chap. 

146. The equation to a curve. All curves of known 
form, such as any of those we have yet considered, have 
characteristic properties capable of definite expression in 
some way or other. Take for instance the hyperbola. 

In the figure a hyperbola is shown in which the asymp- 
totes XX, YY' are at right angles, and therefore called a 
recta?igular hyperbola. Let P be any point on the curve, 
and draw PM, PN parallel to YO, XO. Now take the 
asymptotes as axes of co-ordinates, and denote the abscissa 
and ordinate of P (PN and PM) by x and y, in the usual 
way. It was explained in Art. 105 that one of the properties 
of a hyperbola is that the product of PN and PM is the 
same for all positions of P on the curve ; or, expressed in the 
form of an equation, 

PNx PM= constant ; 
that is, abscissa x ordinate = constant, 

or xy = c . . (1 ). 

This way of stating a property which characterises the 
curve is called the equation to the hyperbola; or, more 
definitely, it is the equation to the hyperbola referred to the 
asymptotes as axes. 

Thus the equation to a curve is the expression of a 
characteristic property in the form of an equation in terms 
of some system of co-ordinates and constants. 

If the reader be making an acquaintance for the first 
time with the idea of an equation to a curve, let him con- 
sider well its meaning. Here, in the case before us, x 
denotes a distance which is altering as the point moves 
along the hyperbola ; so also is y altering ; but their product 
does not alter, it is constant so long as we are considering 
the same curve. 

Whatever curve we may have to deal with, if there be 
some geometrical fact which is known to be true for any and 
every point on it, then the method of representing the 
position of a point by its x and y enables us to express the 
same fact by means of an algebraical equation. It is, so to 
speak, merely a shorthand way of expressing such fact. 



VI 



PLOTTING ON SQUARED PAPER 



155 




Suppose that in the figure PN= 3 and PM= 5, then the 
value of c in the equation to the curve will be 3 x 5, or 15. 
Accordingly, at the point on the curve where the x is 2A, 
the y will be 6 ; at the point where x is 1, y will be 15 ; 
if x = - 5, y = - 3, and so on. 

Thus the equation to any curve involves the idea of a 
point which moves along the curve, and although the 
co-ordinates change their values, yet the two co-ordinates 
are related to each other in a way which does not alter. 

If a different system of co-ordinates be taken, the 
equation to the same curve will be different. Suppose we 
adopt system 2, Art. 143, and define the position of a point 
by its distances from two fixed points. Let the two fixed 
points be the foci F and F' of the hyperbola. A well- 
known fundamental property of the curve is that the 
difference of the focal distances is constant (Art. 105); 

i.e. F'F - FP = constant, or r - r = c, 

where r and r are now the co-ordinates of P in system 2. 



156 PRACTICAL PLANE GEOMETRY chap. 

This, therefore, is an equation to the hyperbola. The 
corresponding equation to the ellipse is 

r + r = c. 

Or take system 3 of Art. 143, and let A (Fig. 143) be 
the focus, and OB the directrix of the hyperbola. Then if 
P be any point in the curve, it is known (see Art. 76) that 
AP: PN= a constant number greater than unity. 

Or writing r and x for the co-ordinates AP and PN in this 

system, the equation is 

r = ex ; 

where c is a constant greater than unity. 

If c be less than unity, the equation represents an 
ellipse. If c = 1, the equation represents a parabola. 

If the student pursues this interesting and important 
branch of mathematics in works on Analytical Geometry, he 
will find that a?iy curve which is of definite form has 
definite equations which represent it. The form of the 
equation depends on the particular system of co-ordinates 
which may be adopted as well as on the curve. 

It is to be understood that in future we shall keep to 
the rectangular system. 

Examples. 1. Find the rectangular equation to a circle, radius 
a, referred to two perpendicular diameters as axes. 

Solution. From any point P on the circle draw perpen- 
diculars PM, PN on the two diameters or axes OX, 
Y, where is the centre of the circle. 
Then, by Euc. I. 47, we have the geometrical property, 
PM* + PN 2 =OP*, 
which expressed algebraically becomes 

2. Using the property of a parabola stated in Theorem 6, Art. 100, 
find the equation to the parabola, taking the axis of the curve 
as the X axis of co-ordinates, and the tangent at the vertex as 
the J' axis. 

Solution. See Fig. 100. The property in question is 
PN 2 = LL 1 xAN, 
or y 2 = /jr, 

where / is the latus rectum LL X of the parabola. 



vi PLOTTING ON SQUARED PAPER 157 

147. To plot a curve, having given its equation. In 

the preceding article it was stated that every known 
curve had some equation which represented it, such equa- 
tion being only a particular way of expressing the law of the 
curve. 

On the other hand, the converse proposition is true, 
namely 

Proposition. Every equation connecting the co-ordinates 
of a point represents a continuous curve or line of some kind. 

We do not attempt to prove this general proposition ; 
our object is to illustrate it by showing how to plot the 
curve in an actual example where the equation is given. 

Let the equation be y o.2t 2 . 

A student acquainted with elementary algebra will recognise in this 
an indeterminate equation ; that is to say, one that cannot be solved in 
the ordinary sense by finding definite values for x and v ; but there 
is an indefinite number of solutions, or pairs of values of x and y 
which satisfy the equation. 

Now the above proposition is equivalent to saying that if all the 
points given by the unlimited number of solutions, or pairs of co-ordin- 
ates x and r, be plotted, such points will not cover the whole plane 
of the paper, but will be confined to definite curves or lines, and will 
occupy these lines entirely, without gaps or breaks of continuity. In 
fact, that the result is the same as if a point were to move in accord- 
ance with the law expressed by the equation, and actually trace the 
continuous curve. 

Let us now test this by actually plotting a number of the solutions, 
and noting the result. 

To plot the curve whose equation is_>' = 0.2 x 2 , we take 
any values for one of the co-ordinates, and calculate the corre- 
sponding values of the other. Thus 

Take x = o, then 7 = o.2(o) 2 = o. 
Again take x = 1, then y = o. 2(1)" = 0.2. 
And take x= 1, thenj = o.2( - i) 2 = 0.2. 



X = 2, 


y = 0.2(2) = 0.0. 


X ' - 2, 


y = 0. 2( - 2)' = 0.8. 


x = 3 


y = 0.2(3)*= 1.8. 


*= -3i 


y = o. 2 \- 3)= 1.8. 


x 4, 


y = 3.2, and so on. 



i 5 S 



PRACTICAL PLANE GEOMETRY 



CHAP. 



Or we might assign any values to y, and then calculate 
the values of x. If this is done it will be convenient to 
transform the equation, and write it 

o.2.v ? =y, or x 2 = $y ; that is, x= *Jsy. 
Now take, say, y - o, then x = V5 x - - 
Take_y=i, then # - V5 = 2.24. 
Take y = - 1, then a- - ^Z - 5 = impossible. 
j/ = 2, x = ^/io = 3.16. 

y = 2, x impossible. 

*= * v/!5 = =*= 3- 8 7- 
x = 4.47, and so on. 



J: 



3 
4> 



Before plotting the points, it will be convenient to 
arrange the results in tabular form, somewhat as follows : 



Table giving Solutions 


OF 


the Equation j = o.2x 2 . 




Values of x 





1 


2 


3 
1.8 


4 




2.24 


3.16 


3.87 
3 


4.47 




Values of.j' 





0.2 


0.8 


3.2 




1 


2 


4 





In plotting these points on squared paper, remember 
that the abscissa x is set off horizontally from O V, to the 
right if positive, to the left if negative ; and that the 
values of the ordinate y are marked off vertically from OX, 
upwards if positive, downwards if negative. 

The result of plotting the above seventeen points is 
seen in the figure. 

Note. The subdivisions of the squared paper are omitted in the 
figure. 

If all possible intermediate pairs of values of x and y 
had been calculated and plotted, the result would have been 
the curve shown in the figure. 

It will be found that the co-ordinates of all points on 
the curve satisfy the given equation, Or are solutions ; and 
that the co-ordinates of any point off the curve do not 
satisfy the equation. That is, the points occupy the curve, 
the whole curve, and no place but the curve. 



VI 



PLOTTING ON SQUARED PAPER 



159 




The curve does not extend below XX, since negative 
values of y lead to impossible values of x. The co-ordinate 
axis O Y is an axis of symmetry of the curve. The upper 
branches extend to infinity, x and y becoming indefinitely 
great together. 

The given equation y = 0.2.V 2 , or x 2 = $y, merely ex- 
presses the law of the curve that one of the co-ordinates is 
proportional to the square of the other. Now referring to 
Art. 100, it is seen that this is one of the distinguishing 
properties of a parabola ; the curve therefore is a parabola. 
On further comparison of the equation with the properties 
stated in Art. 100, it is readily inferred that O Y is the axis 
of the parabola, XX the tangent at the vertex ; that the 
length of the latus rectum is 5 ; and the distance of the 
focus from the vertex O is \ the latus rectum ; that is, 1^. 

Examples. Plot the following curves on squared paper, after 
having first calculated and tabulated the co-ordinates of a 
sufficient number of points on each : 

1. y 2 = o. 2 x z . 5. x = o.2y 2 . 

2. y = o.2(x 3) 2 . 6. y = o.ix 3 . 

3. ><-2=o.2(x-3) 2 . 7. X 2 J = 4. 

4. y = o.2x' 2 1.2X+11. 8. y 2 = Q.^x 3 -\-2.4. 

Note. The first five equations represent the same parabola, but 
differently placed as regards the axes. In 7 the co-ordinate axes 
are asymptotes to the curve. 



i6o 



PRACTICAL PLANE GEOMETRY 



CHAP. 



148. The linear equation Ax + By + C = 0. This equa- 
tion is said to be linear, because it can be proved that it 
always represents a line which is straiglit. It is also said 
to be of the first degree, because it contains no terms, like 
x 2 , xy, y 2 , x 3 . . . ; it involves x and y to the first power 
only. 

The student can easily illustrate by actual plotting, on 
squared paper, that the above equation represents a straight 
line. Take, for example, the linear equation 

2,x - 2y + 6 = o. 

Dividing by 2, and transposing for convenience of calcu- 
lation, we may write it thus 

y-= 1.5* + 3. 

Now give to x any convenient series of values, calculate 
the corresponding values of y, and tabulate the series of 
solutions as described in the last article. Thus 



Table giving Solutions of the Equation y =1.5x4-3. 



Values of x 


n 


- 2 


- 1 


1 


2 


3 




Values of y 


-i-5. 





i-5 


3 4-5 

1 


6 


7-5 





The figure shows these points plotted, with the straight 
line drawn through the points. 

One readily observes from the table or the figure that 
as x increases by 1, y increases by 1^; or thatjy increases 
one and a half times as fast as x. This fact shows the 
locus to be a straight line. 

We give one more example. Let the general linear 
equation Ax + By + C o be transposed into the form 

x y 

- + j= i, 

a 

then a and /> are the distances from O of the points A and 
B in which the line intersects the axes (9 A' and O Y. These 
distances are called the intercepts of the line. Take, for 
instance, the equation previously considered, 



VI 



PLOTTING ON SQUARED I'AI'ER 



161 











Y 

G 
















5 
















4- / 














B 


















3 
Z 
















/ 








X' 


A 












X 





3 / 


? 


i 





1 


z 


j 











-/ 














-z 









Y' 



$x - 2y + 6 = o, 
or 3-v - 2_r = - 6. 

Dividing throughout by - 6, this becomes 



3- v 

-6 

x 



2V 



= I, 



or 



+ ^=i. 



- 2 3 
Now refer to the figure, in which this line is drawn ; it 
will be seen that the intercept a or OA is - 2, and the 
intercept b or OB is 3. 

Examples. Calculate, tabulate, and plot eight points from eacli 
of the following equations. Observe that in each case the 
series of points lie on a straight line. 



1. x=y. 

2. x +y = o. 

3. x+y=i. 

4. x + r +1=0. 

5. y = 2x - 1 

6. y- -\x+ 1. 

7. y = 2x. 



8. y= -%x. 

9. x = 2y. 
10. x y 



3 2 
11. -r + y 

2 7, 



I. 



12. 2. 35.1- + 3. 1 7j'- 4. 86 = 0. 



M 



162 PRACTICAL PLANE GEOMETRY chap. 

149. Problem. To plot a straight line, having given 
its equation, say 4x + 3y - 12 = 0. 

First Method. Select any pair of convenient values of 
x (or y) and calculate the corresponding values ol y (or x); 
plot the points on squared paper ; draw the straight line 
through the two points thus plotted. 

Thus in the given equation 

Put x = o, then y = 4. 

Put x = 4, then 7 = J(i2 - 16) = - 1.33. 

In the figure the points (o, 4) and (4,-1.33) are 
plotted, and the required line is drawn through them. 

Second Method. Transform the equation into the shape 

xy . . 

-- + -=1; the intercepts a and b are then given without 

a b 

further calculation. 

Thus the given equation may be written 

4.V+ &= 12, 

x v 
or - + - = 1 . 

3 4 
The intercepts are now seen to be 3 and 4, and these 
lengths are marked off from O on the diagram, viz. at 
A and B. 

150. Problem. Having given a straight line and the 
axes of co-ordinates (on squared paper), to determine the 
equation to the line. 

First Method. Assume any linear equation for the 
line, say, y = nix + c, where m and c require to be found. 
Read off the values of x and y for any two points on the 
given line. Insert these pairs of values of x and y in 
succession in the given equation. We thus have two 
equations from which to determine /// and c. An illus- 
tration will make this clear. 

Let the given line be PQ in the figure. Select any two 
points on the line, say, P and Q. 

Read off the co-ordinates of P, viz. x = o, y =2.7; and 
the co-ordinates of Q, x = 7, y 4.4. 



VI 



PLOTTING. ON SQUARED PAPER 



163 





Y 










At 

9- 












O 












1 - 


















A 









i 


2 




*X 


-1 
-2- 













149 



5 


Y 
















4. 


\B 












^g, 




3 




















P 
















2 


















1 












\4 











1 


2 


3 


4 


jN 


$ 


7 



150 



Now insert these pairs of values in the assumed equation 
y = mx + c, and we get 

2.7 = o + c, 
4. 4 = ;// x 7 + c. 
From which 

'=2.7, 

AA-c _ 4-4- 2-7 
7 7 



;// 



0-243 ; 



and the required equation to the line is 

y = 0.243.V + 2.7. 

Second Method. If the points where the given line inter- 
sects the axes are available, and not too close together, 
read off the intercepts a and b, and insert them at once 
in the equation, 

x y 

- + j= 1. 

a 

Thus, if the given line were AB (Fig. 150), the inter- 
cept OA or a is 5.6, and the intercept b or OB is 4.7 ; 
therefore the equation to the line AB is 



x y 
5-6 4-7 



1 . 



1 64 PRACTICAL PLANE GEOMETRY chap. 

151. Examples. 1. Plot the following straight lines on squared 
paper. In each case adopt the method of plotting which seems 
the most suitable. 



() y = 3-* - 4- 

(b)y = sx. 
(c) J = 2.r- 3 . 
(d)y=-^x. 


tf) >V = 2. 

('<) jl' = 3- 
(*) x = o. 
(/) j< = o. 


H- 


(w) 2.7x4-3.97' - 6.2 


!-;- 


() 3-5 v - i-U'+ 2 -3 



2. Measure from your diagrams for Ex. I the co-ordinates of the 

points of intersection of the lines c and d ; also of e and_/" ; and 
of in and ;/. In each case verify the result by calculation ; that 
is, by solving the three pairs of simultaneous equations. 
Ain. (1.2, -o.6) ; (4.39, -.488) ; (-0.13,1.69). 

3. Determine the equations to the ten lines given in the figure 

on the opposite page. Answers 

x y , 

1. + = 1. 6. y=i.i2$x. 
4.8 9.1 

2. y=.$x+5.5 7- J= -.83-r. 

3. r = 6.43.1- 23. 1. 8. y \.2.\x 4.6. 

4. .7' = 2.9X + 8. 4. 9. J = X-S-2. 

5. r = .8x + .8 10. j= -2.33x4-2. 

4. Read from the diagram opposite the co-ordinates of the point of 

intersection of lines 1 and 2. Confirm the result by solving the 
simultaneous equations which are given as the answers to 1 and 
2, Ex. 3. Ans. (1.5, 6.25). 

5. Measure, from the diagram opposite, the "slope" of the line 6, 

that is the tangent of the angle which the line makes with OX. 
Find the angle from the table of tangents on p. 20. Am. 6.7 
in 6 or 1. 16 in 1 ; 49. 3 . 

6. Find the slope or gradient of the steepest portion of the road 

shown in Fig. (b), p. 167. Aus. 860 feet in 2 miles ; i.e. 1 
in 12.3, or 0.0813 m L or 4-7- 

7. After having plotted the lines of Ex. 1, measure the angle (1) 

between lines {a) and (6) ; (2) between lines (c) and (d) ; and 
(3) between (e) and (/). Ans. (1) o ; (2) 90 ; (3) 90 . 
Note. Observe (1) that in is the slope of the line y = mx -f- c ; 

(2) that the lines y = nix + c, and y = mx + c f are parallel; 

(3) that the lines y = mx + c andj'= x4V are perpendicular. 



VI 



PLOTTING ON SQUARED PAPER 



165 






















1 




































10 
























9\ 
























81 


'\ / 






















7 


\ 






2^ 




















\ 






3 














/ 6 
























^O 
























4 






\| / 












/ 

1/ 


J 






\ \/y\ i 








1 \ / 1 


2\ 




i/ ' \ 1 

/A \\ 












/ 1 \ 


/ 


V/^' 












-/J j / \/ 












/ 


' 


-/ 




f x ~ 


' / 
-/ 


N 


1 


1 


3 1 


4 \ 


5 


6 




A, 




-9 
























-3 
























\-4 
























\ 




'9 




















~ ,'j 


















5 
imiiii 






6 






\io 


















r 






1 ' 









X 



i66 



PRACTICAL PLANE GEOMETRY 



CHAT. 



152. Plotting the results of observation and experi- 
ment. As will be explained in a future chapter, any 
quantity capable of numerical measurement may be repre- 
sented graphically by a finite straight line set out to a 
specified scale. The co-ordinates of a point, being lengths, 
may be taken to represent two such quantities. We may 
extend this to two quantities of variable magnitude, which 
are mutually related in some manner, and thus exhibit the 
nature of the relationship by means of a curve. From 
many familiar examples we select the following : 

(a) Load-elongaiion diagram. In testing the tensile 
strength of a specimen of mild steel, one square inch 
cross sectional area, if the loads and the corresponding 
elongations are noted at intervals during the operation, 
and subsequently plotted as ordinates and abscissae re- 
spectively, the load-elongation diagram is somewhat like 
Fig. (a), its exact shape depending on the nature of the 
steel. Many testing machines are arranged so that the 
load-elongation diagram may be drawn automatically. 



bnspersq.in. (a) 

30 







\ \ 




\ 








^^~ "v 




<^ s 




L L. 




I 




i 


J/J 




1 

/ft 











elonaati<m 







Z" 



VI 



PLOTTIN T G ON SQUARED TAPER 



167 



(/>) Contour road- map. A cyclist may wish to know 
the elevation above the sea-level, or the gradient at any 
point of his route ; this is well given by a contour road- 
map, Fig. (/>), in which the abscissas are miles travelled, 
and the ordinates are heights in feet above the sea-level. 

LffilaLTTtTiTiYi' i il 



1500 __ 



5 miles 



10 



1500 

feet 
woo 



500 




Sea- 
Level 



(c) Price chart. A manufacturer who studies closely 
the variations in the market prices of materials day by day 
would be interested in diagrams like that of Fig. (c), which 
shows the fluctuations in the price of the metal tin during 
the month of June 1898, the abscissae being market-days, 
and the ordinates the prices in pounds per ton, as quoted 
in the London metal market. 



tper 
71 


to/v 


















W 




















70 












































/ 


69 














/ 




\ 






/ 


\ 


















68 














/ 






\ 


/ 
























67 












1 



































/narked da/us 



i6S PRACTICAL PLANE GEOMETRY chap. 

153. Choice of scales. In plotting curves such as those 
described in the last article, it is important that the scales 
be judiciously chosen. The main points to be attended 
to are now described. 

The horizontal and vertical scales may be chosen quite 
independently of each other, even when both co-ordinates 
represent actual lengths, as in the case of the contour road- 
map of Fig. (/>), page 167. 

The zero points of the scales need not be on the sheet. 
Thus in Fig. (c), page 167, the vertical scale begins at ^67 
per ton, as it would be useless to show anything lower 
than this. Note also that in Fig. 155 the efficiency Scale 
begins at .40. 

The scales should be so chosen that the curve extends ivell 
over the sheet, from bottom to top, and from extreme right 
to extreme left, and is thus not dwarfed either way. 

If the plotted points lie approximately on a straight line, 
the best result is obtained when the line is about equally 
inclined to both axes. See the line in Fig. 155. We should 
avoid adopting scales which would cause the line to be 
inclined at an angle less than say 30 to either axis. 

We may point out some of the bad effects -which result from 
the neglect of these precautions. At the top of Fig. (/'), page 167, 
the contour of the road is drawn in its true horizontal and vertical 
proportions ; so drawn it would not show with sufficient distinctness 
the variations in height and slope of the road, to be of practical use 
to the cyclist. The road would appear to he nearly flat notwith- 
standing that for a couple of miles there is a gradient of 1 in 12, 
almost too steep to be ridden down. In the lower figure the vertical 
heights are magnified twelve times. 

Again, in Fig. 155 at the top, we have set out the efficiency- 
resistance curve with the vertical scale reduced to y^th, to show that 
under these circumstances the curve might easily be mistaken for an 
approximate straight line. 

Or again, in Fig. (<?), page 166, the Inst portion of the diagram 
appears to be a perfectly straight line, nearly vertical ; in order to 
know whether the line is actually straight or not, it would be neces- 
sary to enlarge the horizontal scale for this portion of the figure. 
The enlargement would require to be from fifty to one hundred times, 
in order to comply with the condition stated above. 



VI 



PLOTTING OX SQUARED PAPER 



169 











































J 



































^ 


^ 






















V 
















































! 








/" 


















































1 

/ 


-3 







6 






3 







6 


0" 


9 





A 


V 


































/ 










-1" 














:: 



























154. Interpolation. Maxima and minima. When we 
are given the values of a quantity for a series of intervals, 
we can readily find any intermediate value by plotting the 
given series. This is called interpolation. VVe can also 
determine in a similar manner the maximum or minimum 
value of a quantity, where such exists, if we are given a set 
of values of the quantity ranging on each side of the 
maximum or minimum. The following example should 
make this clear : 

Example. In a steam engine the displacements of the slide valve 
from its mean position, corresponding to a series of angular positions 
of the crank, are given in the following table : determine 

(a) the displacement of the valve when the crank angle is io ; 

(/>) the angle of the crank when the valve displacement is zero ; 

(<) the maximum displacement of the valve. 



Angle of crank . 


-6o 


-3 


o 


3o 


6o 
2-95" 


90 


Displacement of valve 


-.91" 


.80" 


2.17" 


2-93" 


2.23" 



In the figure above these positions are shown plotted on squared 
paper, with crank angles as abscissae, and valve displacements as 
ordinates. & fair curve is drawn freehand through the six points thus 
obtained. The required results are then read off from this curve. 
Thus we obtain : Ans. (a) 2.49" ; {/>) -43.8 ; (c) 3.06". 



170 



PRACTICAL PLANE GEOMETRY 



CHAP. 



155. Laboratory test of a crane. The following numbers 
were obtained by testing a small crane under different 
loads : 



Load raised by 
crane, in lbs. or 
Resistance, /_ 


o 


1 

20 | 40 


60 


80 


100 


120 
_____ 

38.1 


140 


1 


Force exerted at 
handle, in lbs. or 
Effort, E . 


4-5 


I0.7 1 15.6 

1 


21.7 


27.0 


32.8 


43.6 


Efficiency, / (cal- 
culated) 


o 


i 

47 i -64 

1 


7i 


74 


,6 


79 


.80 



The gearing of the crane was such lhat, but for friction, the force 
required at the handle would have been \ the load raised. The 
efficiency is calculated by dividing this force (without friction) by the 
actual force. Thus when So lbs. is being raised, the effort required if 
there were no friction would be 20 lbs., whereas the actual effort from 
the table is 27 lbs. The efficiency for this load is therefore 20-^27 
= .74, or 74 per cent. 

Now, plotting the efforts and resistances as co-ordinates 
on squared paper to suitable scales, we obtain points which 
are seen to lie very nearly on a straight line. 

The best average position amongst the points for this line is readily 
judged by applying to the diagram a piece of black thread, or a piece 
of tracing-paper on which a straight line has been ruled. 

The efficiency-resistance curve is shown in the same 
figure. The points on it are plotted, and a fair curve 
drawn so as to lie evenly amongst them, so far as can be 
judged by sight. This curve may be used to correct 
errors of observation. 

The two curves may be said to be set out on a resistance 
base ; the horizontal scale of resistance is marked in the 
figure. The efforts and efficiencies are plotted as ordinates, 
their respective scales being shown on the left and right of 
the diagram. 



VI 



PLOTTING ON SQUARED PAPER 



171 






100 



50 







Elbs 




O ZQ 40 60 80 100 120 R 140 H>S. 

155 



]'/2 



PRACTICAL PLANE GEOMETRY 



THAT. 



158. Examples on Plotting. 

1. Plot the following values of /, the pressure in pounds per 
square inch, and t the temperature Fahrenheit, of dry saturated 
steam ; join the points by a fair curve, and read off the value 
of/ when t is 293 . Ans. 60.4 lbs. per sq. in. 



\ 

t 


212 23O 


248 


266 


284 


302 


320 ! 338 ; 356 374 


392 


p 


14.7 j 20.b 

1 


28.8 


39- 2 


52-5 


69.2 


J 

89.7 1 15. 1 j .45.8 182.4 

1 ! 1 


225.9 



2. The values of A, the average attendance (in thousands) at the 
primary evening schools in England for the years 1S87 to 1S97 
are given ; obtain a curve showing these values. 



Year 


1887 


1888 


1889 


1890 


1S31 


1892 


1893 


1 '-'94 


1895 


1S96 1897 


A 


30.6 


33-3 


37-i 


43-3 


52.0 


65.6 


81. 1 


"5-5 


129.5 


147 


179.6 



3. Find one root of the cubic equation 2.v 3 - $x - 16 = 0. 

Method. Let y=2x 3 - 3-r 16. Give to x the values 1,2, 
3, etc., and calculate the corresponding values of_j'. Plot these 
values of x and y on squared paper and draw a fair curve 
through the points so obtained. Read off the values of x where 
y is zero, that is, for which 2x 3 ix - 16 zero. 

To obtain a more correct solution, after observing that one 
required value of .v lies between 2 and 3, take closer values of 
x, say 2.1, 2.2, 2.3, etc., and after calculating the values of y, 
plot these new values of x and_y to a bigger scale. Ans. One 
solution is .1=2.25. 

N.B. In this manner one or more solutions to any equation 
may be obtained. 

4. The half-ordinates of the load water-plane of a vessel are 12 

feet apart, and their lengths are 0.5, 3.8, 7.7, 1 1.5, 14.6, 
16.6, 17.8, 18.3, 18.5, 18.4, 18.2, 17.9, 17.2, 15.9, 13.4, 
9.2, and 0.5 feet respectively. Determine (1) the total area 
of the plane in square feet, (2) the position of the centre of 
area of the section of the vessel made by the water-plane. 
Ans. (1) 2649 sq. feet : (2) 102.8 ft. from the first ordinate. 

Hint. (1) Find the mean ordinate as in Art. 43, and 
multiply it by the total length to obtain the area. (2) Multiply 
each mid-ordinate by its distance from the first ordinate, and add 
these products together ; divide the result by the sum of all 
the mid-ordinates. 



vi PLOTTING ON SQUARED PAPER 173 

157. Approximate linear laws. In Art. 155, referring 
to the crane, it was seen that when the force applied to 
work the crane, and the load raised that is, the effort and 
resistance were plotted as co-ordinates on squared paper, 
the points obtained lay very nearly on a straight line ; 
the law connecting the two is therefore said to be linear, 
or approximately linear. The determination of the laiv of 
the crane is the same as that of finding the equation to this 
straight line. The line was located by means of a stretched 
thread or ruled tracing-paper. 

We proceed exactly as in Prob. 150. Let the equation 
to the line be 

y = ax + b, 

or E = aR + />, 

since the ordinate y is the effort E, and the abscissa x is 
the resistance R. 

To find the constants a and /', select any two points on 
the line, say P and Q, near the ends. Read off the 
co-ordinates of R and Q. Thus 

for R, E = 4.gi lbs., R = o ; 
and for Q, ^ = 43.8 lbs., R= 140 lbs. 

Substitute these values in the above equation, and we 
obtain 

4.91 lbs. =0 + b, 
43.8 lbs. ax 140 lbs. + b ; 
from which 

b = 4.91 lbs. 

43.8 lbs. - 4.91 lbs. 
a = - = .278. 

140 lbs. 

Therefore the equation to the effort-resistance line, or 
the law of the crane, is 

. = .278^+4.91 lbs.; 

Or, the force that must be applied at the handle in 
order to raise any load is equal to 4.91 lbs. plus .278 
of the load. 



174 



PRACTICAL PLANE GEOMETRY 



chap 



158. Miscellaneous Examples. 

1. In a Stephenson link motion the following measurements 
were made : - 



Angle of crank . 


-45 


-i5 


i5 


45 


75 


105' 


Displacement of valve 


-25" 


.98" 


1.86" 


2.24" 


2.05" 


1.32" 



Determine (a) the displacement of the valve when the crank angle 
is zero ; (b) the angle of the crank when the valve displacement 
is zero ; and (<') the maximum displacement of the valve. 
Ans. (a) 1.47"; (b) -39.2 ; (c) 2.25". 

2. The following numbers refer to the test of a crane, determine 
the linear law of the crane. 



Effort E, lbs. 


7 
4 


36 


21 

65 


28 
93 


35 


4-' 


49 


56 


63 


Resistance R, lbs. 


122 


151 


183 


2l6 


248 



Ans. E = o. 23 A'4-6.54 lbs. 
3. Try if an equation of the form xy = ax-\-by approximately repre- 
sents the relation between x and y, pairs of values of which are 
given in the table below, and if so, determine the mean values 
of a and b. 







, 












Values of j' 


5 


6 

28 


7 


8 


9 


10 


1 1 


Values of x 


18 


54 


133 


-455 


- 1 1 1 


-65 



















Ans. a = 8. 7, b = - 13, or xy = 8.jxy 13. 
Hint. The equations xy = a x + by may be written in the form of 

1 = a- + b-. Therefore calculate and plot the values of the 
y x 

reciprocals - and - as co-ordinates. If these are called Fand 
y x 

X, we have then to determine a and b in the approximate linear 

law (if such exists) 1= aY+bX, as in Arts. 150 or 156. 



VI 



PLOTTING ON SQUARED PAPER 



1/5 



A series of pressures p and volumes v of saturated steam, as 
determined experimentally, are given in the table below. The 
logarithms of these quantities are also given. By plotting log/ 
and log v, try whether a relation of the form log/ + <z log vb 
(i.e. pv a = cox\sL) holds approximately between them, and if 
this is found to be the case, determine the best average values 
for the constants a and /'. 

Properties of Saturated Steam 



Pressure, /. Lbs. 
per sq. inch 

Volume, v. Cubic 

feet per lb. 


I.o6 


2.88 

122 

.462 
2.09 


6.86 


14.7 

26.4 
1. 17 
1.42 


28.8 

14.0 
1.46 


52-5 
7-97 


89.7 
4.82 


146 
3.06 


226 
2.02 


313 


53-4 
.837 


log/ . 


.020 
2.50 


1.72 


i-95 


2.17 
483 


2-35 


log 7' . 


i-73 


i-'5 


.902 


.683 


35 



Ans. (1=1-0$; 3 = 2.66 (or / 1,05 = 46o). 
5. A log of timber 20 feet long has the following cross-section 
areas at the given distances from one end. Find the volume 
in cubic feet. 



Distance from one end in feet 





2.6 


5 


7.8 


12 

3-5 


15 


17.6 

3-i 


20 


1 
Area in square feet . . ' 5.0 

1 


4-3 


3-8 


3-6 


3-3 


3- 



A?is. 72. 7 cub. ft. 
Hint. It will be observed that the distances between the given 
sections are unequal. In such a case first plot the given 
numbers on squared paper, and then divide into, say, 10 strips 
of equal width. Find the mean cross-sectional area by the 
mid -ordinate method, and multiply by the total length of 
the log. 



SFXTION II 

PRACTICAL SOLID GEOMETRY, OR 
DESCRIPTIVE GEOMETRY 

CHAPTER VII 

POSITION IN SPACE DEFINED AND EXHIBITED 

159. Introduction. Hitherto the points, lines, and 
figures in the various problems taken have been confined 
to one plane, represented by the plane of the paper ; we 
now pass on to the more general case and treat of the 
geometry of space. Our diagrams must now represent 
the three dimensions of length, breadth, and thickness, 
whereas previously we were concerned with only two, 
length and breadth. 

It is necessary to distinguish between pure and practical 
solid geometry. 

Pure solid geometry deals with the geometrical relations 
which exist amongst points, lines, and surfaces in space. 
Practical solid geometry shows how to exhibit these rela- 
tions by scale drawings which can be measured. We have 
an illustration of the former in the eleventh book of Euclid, 
where a number of definitions relating to solid figures are 



vii POSITION IN SrACE DEFINED AND EXHIBITED 177 

given, followed by the propositions, arranged in strict 
logical sequence and proved. The diagrams are of 
secondary importance, being generally semi -perspective 
sketches, just sufficient to convey an idea of the form of 
the solid figure. It is necessary that the student be 
acquainted with the definitions and some of the proposi- 
tions of Euclid XL, and for convenience, these are given 
in an appendix at the end of this volume. 

In practical solid geometry we are concerned not so 
much with the proofs of propositions, as with the methods 
whereby the relative positions and forms of figures in space 
of three dimensions can be exhibited on a surface such as 
a sheet of drawing-paper, which has only two dimensions, 
and in such a manner that these positions and forms can 
be ascertained from the diagrams or drawings by direct 
measurements to scale. 

As in the preceding section, we shall endeavour to 
enforce upon the student the importance of good draughts- 
manship. A leading idea throughout should be that in 
making practical computations, graphical processes may 
with advantage often be employed in preference to other 
methods. One is only properly equipped for the work when 
accuracy in execution has become habitual. 

The special difficulty which the beginner experiences is 
to be able to conceive clearly the connection between the 
figures drawn on paper and the actual figures in space 
which they represent. This difficulty is greatly lessened 
in the initial stages by making free use of models. The 
problems of this introductory chapter are arranged so 
that the student may easily make his own models for all of 
them. It has been found that by their use very clear 
and precise notions are formed at the start, and an 
excellent method of working is introduced. 

The problems relate to ways of exhibiting the positions 
of points, lines, and planes in space, and of finding the 
distances and angles between them. They form the basis 
of the method by which solid form is defined by drawing. 

N 



178 PRACTICAL SOLID GEOMETRY chap. 

160. The position of a- point in space denned by rect- 
angular co-ordinates. We have seen in Art. 144 how the 
position of a point in a plane may be defined by reference 
to two perpendicular axes. In like manner the position of 
a point in space may be defined by reference to three 
mutually perpendicular planes. For example, the situation 
of a small object in a room is known if we know its height 
above the floor and its distance from each of two adjacent 
walls. 

Let the planes of 'reference be as shown in the figure, one 
horizontal and the other two vertical, and at right angles to 
each other, the latter being distinguished as the front and 
side vertical planes. These planes intersect in three lines, 
also mutually perpendicular, called the co-ordinate axes, one 
being vertical and the other two horizontal. The point 
common to the three axes and the three planes is called 
the origin, and is denoted by the letter O, the three axes 
being labelled respectively OX, OY, OZ, the latter being 
the vertical one. The three planes of reference may 
also be designated as the planes of X Y, YZ, and . ZX 
respectively. 

The position of any point A may now be defined by 
its perpendicular distances Ad, Ad' , Aa from the three 
planes of reference. These three distances are called the 
rectangular co-ordinates of the point, and may be denoted by 
x, y, z. The co-ordinate x is the distance of A from the 
plane of YZ, measured parallel to the axis OX. Similarly 
for the other two: y is measured parallel to OY, and z 
parallel to OZ. Thus x and y are the horizontal co- 
ordinates, and z is the vertical co-ordinate. 

A co-ordinate has a negative value when the point is 
situated on the other side of the plane from which that 
co-ordinate is measured. Thus z would be negative for any 
point below the horizontal plane XY. And x would be 
negative for any point at the back of the front vertical 
plane YZ. Similarly, y would be negative for any point to 
the left of the plane XZ. 



vii POSITION IN SPACE DEFINED AND EXHIBITED 179 




The student will note that the co-ordinate planes, when 
extended eacli way from 0, divide the neighbouring space 
into eight trihedral angles. He should also note that when 
the co-ordinates of a point are given in sign and magnitude, 
the signs tell us in which trihedral angle the point is 
situated, and the magnitudes give lis the exact situation in 
this angle. Thus the position of any point in space is 
completely defined, without ambiguity, by its three rect- 
angular co-ordinates. 

The point whose co-ordinates are x, y, z may for short- 
ness be referred to as the point (x, y, z). 



Examples. State in which trihedral angles the following points 
are situated : 

1. The point ( - I, 1, 1). 

Ans. Behind YZ, to the right of ZX, above XY, or in the 
back-right-upper angle. 

2. The point (1, - 1, 1). 

Ans. The front-left-upper angle. 

3. The point (1, - r, - 1). 

Ans. The front-left-lower angle. 

4. The point ( - I, -I, - I). 

Ans. The back-left-lower angle. 



i8o PRACTICAL SOLID GEOMETRY chap. 



161. The position of a point in space exhibited by 
projection. In descriptive geometry it is not enough to 
define the position of a point in space. It is further neces- 
sary to set out this position by a scale drawing. 

As the drawing is necessarily confined to one plane, and 
the figure which the drawing represents is not plane, some 
convention is required in order to connect the two, and 
so render the signification of the drawing precise and 
definite. 

To illustrate, let the student take a quarter of an 
imperial sheet of drawing-paper, and on it rule two perpen- 
dicular lines, distant about 5" from the upper and left 
edges, as shown in Fig. 161 (a) at ZY, ZX. Then cut the 
paper along the horizontal line from O to Z, and indent ! it with 
a blunt instrument from O to Kand Z to X. Now fold the 
paper along the indented lines, and secure the overlapping 
parts on the left by means of a paper-fastener. A partial 
model of the planes of reference, Fig. 1 6 1(/>), is thus obtained, 
comprising the front-right-upper trihedral angle, the co-ordi- 
nates of points situated in which are all positive. 

Suppose now it is required to set out the position of a 
point A, whose co-ordinates x, y, z are given. 

Unfold the model, and along OX and OY mark off 01 
and Om equal to x and y, and along OZ, OZ set off On, 
On, each equal to z. Through /, m, and n draw the lines 
parallel to the axes, as shown, intersecting respectively in 
a, a , a". Then this plane figure, or unfolded model, with 
its lines and points, is the scale drawing or diagram which, 
properly interpreted, represents the position of the point A 
as regards the three planes of reference. 

If this figure were given, then in order to determine 
from it the x co-ordinate of the point A, we should measure 
the length of 01, ma, or na" ; to ascertain the y co-ordinate, 
we should measure Om, la, or na' ; and for the z co-ordi- 
nate we should measure any one of On, On, ma', &&a". 

If it were desired to show exactly what the drawing 
represented, we should fold it along the lines of the axes, 



vii POSITION IN SPACE DEFINED AND EXHIBITED 181 



161(a) 





z 


f 








n 


i a' 




/ 

f 




\m 


z 


n 







CV" 


I 


>a 



Y 



X 



161(b) 




7^7y 



so as to obtain the model of the planes of projection ; and 
to show the actual point A in space and the three per- 
pendiculars from it, we might use a small sphere with three 
projecting pieces of wire, Aa, Aa, An". 

The points a, a, and a" are called the three projections 
of the point A. But some formal definitions are required, 
and we give these in the next article. 



1 82 PRACTICAL SOLID GEOMETRY chap. 



162. Definitions relating to projection. If from the 
various points which may be supposed to constitute any 
object, straight lines proceed to meet a plane, the object 
is said to be projected on the plane. 

The straight lines are called projectors. 

Parallel projection is the case in which the projectors 
are all parallel to one another. 

Radial or perspective projection that in which their direc- 
tions all pass through one point. 

Parallel projection is subdivided .into oblique and 
orthogonal or orthographic ; in the former the projectors 
"are inclined to plane of projection, in the latter they are 
perpendicular. When the word "projection" is used with- 
out qualification, orthographic projection is to be understood. 

Thus the projection of a point on a plane is the foot of 
the perpendicular let fall from the point to the plane. 

Referring again to Figs. 1 6 1 (a), 1 6 1 (/>), the points a, a, a" 
are seen to be the three projections of A on the planes of 
reference. The projection a on the horizontal plane is called 
the plan ; the two on the vertical planes are called elevations : 
a is the front elevation, and a" the side elevation. In 
Fig. 161 (a) the points is said to be represented by its 
projections. The planes of reference are also called planes 
of projection. 

The lines aa and aa" in Fig. 1 6 1 (a) are also called pro- 
jectors. They are perpendicular respectively to O Fand OX. 

We may point out that in all cases the three projections show which 
trihedral angle contains the point. Thus in Fig. 162 (,/) the point A 
is behind the plane of YZ, making the x co-ordinate negative. 
The point is projected on the planes of reference at a, a', a". Now in 
order to derive Fig. 162 (b), the vertical planes of Fig. 162 (a) must be 
rotated about the axes of A" and Y respectively, and always in the 
same directions. These directions of rotation are those given when 
describing the construction of the model, and are such as to open out 
the positive dihedral angle. The angle in which the point lies and 
its exact situation are given without ambiguity by Fig. 162 (b). 

Notation. A capital letter is used to denote the point in space, 
and the corresponding small letters denote the projections. Thus for 
the point A the plan is a, the front elevation a , and side elevation a". 



vii POSITION IN SPACE DEFINED AND EXHIBITED 1C3 




Z,X' 



a; 



!/ 



n 



\a'. 







m. 



162(a) 



X 162(b) 



ZX 



Examples. 1. The x, y, 2 co-ordinates of a point A are 
respectively 3", 4", and 2" ; draw the three projections of the 
point and set up the model for this case. 

2. Draw the projections of the three points (2.5", 3", 4"), ( 3", 

4", 2"), and (3", - 4 ", -2"). 

3. In which trihedral angle is the point A situated whose projec- 

tions are given in the figure below ? Measure the co-ordinates of 
the point, scale th. Ans. -1.6", -3.6", -3". 

Note. OX', OY', OZ', 
OZ' represent the 
negative directions of 
the axes. 01, 0/n, 
On are the x, y, z 
co-ordinates of A, 
and are seen to be all 
set off in the negative 
directions along their 
respective axes, from 
which we infer that 
the three co-ordinates 
are all negative. 

Or thus Set up the model and in some way represent the 
point A in its actual position, say by using a lady's hat-pin. 

Look vertically downwards on the model ; the plane ZOY 
appears as the line OY, and because the //an a is behind Y'OY 
the point A must be behind the plane ZOY. Also the plane 
ZOX appears as the line OX, and since the plan a is to the left 
of OX, the point A is to the left of the plane ZOX. Now view 
the model in the direction at right angles to the plane ZOY; 
the plane XOY appears as the line OY \ and because the front 
elevation d is below OY the point A is below the plane XOY. 
Hence the point A as represented by its projections a, a', a" 
is in the back-left- lower trihedral angle, and its co-ordinates are 
found on measurement to be as given above. 









I 


,a" 




y 






Y 


z 


a'. 




71 


." 


}ro 


Z' 











z;x 



1 84 PRACTICAL SOLID GEOMETRY chap. 

163. Problems on the position of a point. We suggest 
here some simple problems, and give some examples for 
the student to work out himself. It is intended that the 
drawing-paper shall be cut, indented, and folded as 
explained in Art. 161, and then fixed to the board with the 
vertical planes YZ, ZX free to be laid flat while drawing, 
or lifted into position while studying the problems, as often 
as may be required. All the problems reduce to one or 
other of the solutions of the right-angled triangle specified 
in Art. i 2. 

We give only hints. If formal solutions were written we 
should defeat the object in view, which is to familiarise 
the beginner with the method and signification of projections, 
by encouraging the use of models. 

From the lower figure it is seen that the point A and 
its projections a, a', a", the points /, m, n, and O form the 
eight corners of a rectangular prism, and OA is a diagonal 
of the solid. Diagonals of some of the faces of the prism are 
shown by dotted lines. We now suggest some problems, 
having given as data the co-ordinates of the point A. 

(a) To find the distances of A from the axes. 

The distance of A from OX is equal to .41 or Oa ', that is to the 
hypothenuse of the right-angled triangle Oma', the two sides of which 
are two of the given co-ordinates. 

(b) To find the distance of A from the origin. 

Here we must find the length of the hypothenuse OA of the right- 
angled triangle Oa'A, one side Ad being the given x co-ordinate, and 
the other side Od being known from the above. 

(c) To find the angles which OA makes with the -planes 
of reference. (See Def. 8, Appendix II.) 

The inclination of OA to the horizontal plane is the base angle 
AOa of the right-angled triangle AOa. The inclination of OA to the 
front, vertical plane YZ is the base angle AOa of the right-angled 
triangle AOa'. 

(d) To find the angles which OA makes with the axes. 
The inclination of OA to the vertical axis OZ is the vertical angle 

A On of the right-angled triangle A On. This angle is the complement 
of the angle AOa. 

(e) To draw the three projections of OA. 
These are the lines Oa, Od, and Oa". 



vii POSITION IN SPACE DEFINED AND EXHIBITED 1S5 



7. 



163(b) 



163(a) 








ft 


a' 






/ 

/ 






/ 
; 




77b 




z 


ru 







Y 






I 


X 




c 


v * 


i 




X 







Examules. 1. The x, v, s co-ordinates of a point A are respec- 
tively 3", 4", and 2".' 

(a) Draw the three projections of A, viz. a, a', and a". 

(b) Draw the three projections of OA, and measure their 

lengths. A)is. 4.47", 3.O1", 5" on the planes of YZ, 
ZX, X V respectively. 
(C) Determine and measure the distances of A from the axes 
of A', J', and /.. Ans. 4.47", 3.61", 5.0". . 



186 PRACTICAL SOLID GEOMETRY chap. 

(d) Determine and measure the distanceof A from the origin O. 

A us. 5.39". 

(e) Determine and measure the angles which OA makes 

with the planes of projection. A /is. ^j.S , 48. o, 
21.8 with the planes of YZ, ZX, X Y. 

(f) Determine and measure the angles which OA makes with 

the axes of A', J', and Z. Ans. 56. 2, 42.0, 68.2. 

(g) Determine and measure the true distances of the projec- 

tions a, a, a" from one another. A us. a" a = 4.47" ; 
fl a' = 3.6i"; dd'=$". 
(h) Set out to scale the true shapes of the four triangles 
aa'a", Oa"a, Oaa', Odd'. 
Note. In g and h the word true signifies that the results are to be 
obtained for the true positions of the planes of projection, and 
not for their positions when laid flat. 

2. The three projections a, d, a" of a point A are given in Fig. 

163 (a). Copy this figure double size, then measure the co- 
ordinates, and obtain the results c to h of Ex. 1. 

3. Draw the figure of Ex. 3, Art. 162, full size, then obtain the 

results of Ex. 1 above. 

164. Polar co-ordinates of a point. The position of a 
point in space might be defined in other ways, e.g, by its dis- 
tances from three fixed points. But the only two systems of 
co-ordinates in general use are the rectangular and the polar. 

In the polar method we choose a point called the 
pole, an axis OZ through the pole, and a plane ZOX con- 
taining the axis. The polar co-ordinates, say (r, 6, 4>) of 
any point A are then (1) the polar distance OA or r ; (2) 
the angle ZOA or 6 ; and (3) the angle between the planes 
of ZOA and ZOX, say <. 

In illustration, consider how a place on the earth's 
surface is located. In the figure is the centre of the 
earth, OZ the north polar axis, DXE the equator, ZOX the 
meridian plane through Greenwich, intersecting the surface 
in the meridian circle ZGX and the plane of the equator 
in OX. Let A be the place on the surface, and let the 
meridian plane ZOA intersect the surface in the meridian 
circle ZAM, and the plane of the equator in OM. Then 
the position of A is usually defined by its longitude, that is, 
the angle XOM, and its latitude MO A. 



vil POSITION IN SPACE DEFINED AND EXHIBITED 187 




Choosing OZX for reference, the polar co-ordinates 
(r, d, 4>) of the point A would be respectively the earth's 
radius OA, the aTlatitude ZOA {i.e. 90 - latitude), and the 
longitude XOM. 

In the figure the polar co-ordinates of A are rOA, 
6 = angle ZOA, and </> = angle XOM. 

The problems resolve as before into cases of the 
solution of the right-angled triangle. 

Examples. 1. The rectangular co-ordinates of a point A are 3", 
4", and 2" ; find the polar co-ordinates (r, d, 0). 
Ans. 5.39", 6S.2 , 53. i. 

2. The polar co-ordinates ;-, d, <p of a point A are respectively 3", 40, 

and 55. Draw the projections of the point, and measure its 
rectangular co-ordinates. Ans. 1.11", 1. 58", 2.3". 

3. Find the polar co-ordinates of the point A, represented in projec- 

tion in Fig. 161 (a), drawn | size. Ans. 5.46", 70.7 , 66. 5 . 

4. Taking the latitude and longitude of Rome as 41. 9 N. and 

1 2. 5 E., represent the position of Rome by a plan and two ele- 
vations. Radius of the earth 4000 miles. Scale i"to 1000 miles. 

5. A line 3-i" long from the origin to a point A makes angles of 

40 and 6o respectively with the axes of X and Y. Draw the 
projections of the line, and measure the rectangular and polar 
co-ordinates of the point A. Ans. 2.6S", 1. 75", 1.4 1" ; 
3.5", 66.2% 33. i. 



iSS PRACTICAL SOLID GEOMETRY chap. 



165. The straight line. In Fig. 165 (/>), let AB be a 
straight line in space, and ab, a'b', a'b" its three projec- 
tions. Then these projections, given as in Fig. 165 (a), 
completely represent and define the line. 

Or these projections could be drawn, if as data we were 
given the co-ordinates of the two ends of the line. 

Again arrange the drawing-paper so that the planes YZ, ZX 
may be lifted into the vertical position at any time. The line AB 
in space may be exhibited in the model by cutting a piece of paper 
to the shape ABba and attaching it to the plane XY, as shown by a 
folded margin left for the purpose ; in setting out this shape observe 
that ab is equal in length to the plan of the line, and the perpendiculars 
a A, bB are equal to the z co-ordinates. 

Examples. 1. The rectangular co-ordinates of two points A and B 
are respectively 2", l", {", and \\ 3", if". 

(a) Draw the three projections of AB, and measure their 

lengths. Ans. 2.24", 1.8", 2.5", on YZ, ZX, XY. 

(b) Determine and measure the actual length of AB. 
Ans. 2.69". 

That is, find the length of the hypothenuse of a right- 
angled triangle, of which the base is equal to the plan 
ab, and the height to the difference of the 2 co-ordinates. 
(C) Determine and measure the angles which AB makes 
with the planes YZ, ZX, XY. Ans. 33. 9 , 48, 21. 8. 

That is, find the angles between AB and its three pro- 
jections a'b', a"b", ab. These all reduce to finding the 
base angles of right-angled triangles. 

(d) Determine and measure the angles which AB makes with 

the axes of X, Y, Z. Ans. 56. i, 42, 68.2. 
Since Aa is parallel to OZ, the angle which AB makes 
with OZ is equal to the angle aAB. Similarly for the 
other two angles. These angles are determined as the 
vertical angles of right-angled triangles, and are com- 
plementary to the angles of (c) above. 

(e) Determine the projections of the points where AB pro- 

duced meets the planes of projection YZ, ZX, X Y. 
Measure the co-ordinates of the three traces so found. 
Ans. (o, 3.67", 1.83"), (2.75", o, o), (2.75", o, o). 

(f) Determine the true shape of the triangle OAB, and 

measure the angle A OB. Ans. 72. 8. 
First find the lengths of the sides, and then construct 
the triangle. 



vii POSITION IN SPACE DEFINED AND EXHIBITED iSg 







z 














n 


at 


lesta^ 


4 

i 

i 


/ 


f 


r 






b' 










\n 


r 







rrv 




Y 


Z ,j 

OL L. 


L 




a, 




9 


>c " 


y 






-^^^ ^"-^ 


^"^^. ^^v 


b 


P 














X 










Copy Fig. 165 (a) double size. Then measure the co-ordinates 
of A and B, and obtain the results b to /of Ex. 1. 

Find the length of the line which joins two opposite corners of 
a building brick 9" x 45" X 3". And determine the angles which 
this line makes with the edges and faces of the solid. Ans. 
Length =lo|". Angles with the 9", 4V', 3" edges = 31.0', 
64.7 s , 73-4 Angles with the small, middle, and large faces 
= 59.0, 25.3', 16.6 . 



190 PRACTICAL SOLID GEOMETRY chap. 



166. The plane. A plane, of indefinite extent, is located 
when the positions of any three points in it, not in one 
straight line, are known. In the rectangular system, the 
points A, B, C, where the plane meets the axes, are the 
three most convenient points to choose for the purpose 
of defining the position of the plane. The lengths OA, 
OB, OC are called the intercepts of the plane on the 
axes. We may denote the lengths of these intercepts by 
a, b, and c. The lines BC, CA, AB, where the plane 
intersects the planes of projection, are called the traces of 
the plane, and serve very well to represent the plane by 
projection. 

Thus, a plane is very cox\\zme\\\\y defined in the rectangular 
system by its intercepts, a, b, c ; and represented in projectio?i 
by its traces, BC, CA, AB 

To the data of Ex. I below, let the student make the following 
model. Cut out in paper a triangle whose sides are equal to AB, BC, 
CA, Fig. 1 66 (a), leaving a folding margin along AB for attachment 
to the horizontal plane as shown in Fig. 166 {/>) ; on it draw CD per- 
pendicular to AB. Next draw OD perpendicular to AB and cut out 
in paper a right-angled triangle, having OD for base, and OC for 
height, leaving a margin along OD to be attached to the horizontal 
plane, underneath the plane ABC, as shown in the lower figure. The 
first example should now be worked by the student himself, without 
assistance other than that given in the notes. 

Examples. 1. The intercepts a, b, c of a plane on the axes of 
X, Y, and Z are respectively 4", 5", and 3". 

(a) Draw the three traces of the plane AB, BC, CA, and 

measure their lengths. 
Ans. 6.4", 5.83", 5"- 

(b) Find the rabatments of the triangle ABC into the three 

planes of projection. 
A plane is said to be rabatted when it is turned about its 
trace into the plane of projection. Thus ABC is rabatted 
into the horizontal plane by rotation about AB ; and 
into the planes of YZ, ZX, by being turned about BC 
and CA respectively. 

(C) Determine the rabatments of the triangle COD into each 
of the three planes of projection. 



VII POSITION IN SPACE DEFINED AND EXHIBITED ini 



z 






/ 


c 


166(a) 


/ 
/ 

/ 
/ 

/ 

1 

I 






c\ 




^\-# 


z \ 


\ \ 


jr+ys 


X 


_/i/e/' 





166(b) 




y 



(d) Determine and measure the angles which the plane ABC 
makes with the planes of YZ, ZX, XY. 

Ans. 5 7; 2, 64. 4 , 43. S. 

The inclination of the plane ABC to the horizontal plane 
is measured hy the angle ODC o{ the right-angled triangle 
ODC. Corresponding constructions are required for 
the inclinations to the other two planes of projection. 
See Def. 9, Appendix II. 



192 PRACTICAL SOLID GEOMETRY chap, vii 



(e) Determine and measure the angles which the plane makes 

with the axes of X, Land Z. 
Ans. 32.8 , 25-6, 46. 2 . 
These angles are complementary to the angles of (e). 

(f) Draw the three projections of OP, the perpendicular, 

from the origin to the plane. Determine and measure 

the true length of this perpendicular. 
Ans. 2.17". 
The point P is in CD, and OP is perpendicular to CD. 

(g) Find and measure the angles which the perpendicular 

OP makes with the planes YZ, ZX, XV. 
Ans. 32. 8, 25. 6, 46. 2. 

(h) Find and measure the angles which OP makes with the 
axes of X, Y, and Z. 
Ans. 57.2 , 64.4 , 43.S . 

Note the illustrations of the theorem, that the angle 
between two planes is equal to the angle between any 
two lines which are perpendicular to the planes. 

2. Copy Fig. 166 {a) accurately, double size ; then measure the 

intercepts and obtain the results b to g of Ex. j. 

3. The latitude and longitude of a place A are 52 N. ; and 

those of a place B are o and 40 W. (a) Find the angle sub- 
tended by the two places at the centre of the earth. (/>) Find 
the length of the shortest path on the earth's surface between 
the two places. Radius of earth = 4000 miles, (c) Find the 
direction of this path at each place. Ans. (a) 62. 4 . (b) 4410 
miles, (i) At A, 46. 5 W. of S. ; at B 28 E. of N. 

4. The latitude and longitude of Rome are respectively 41.9 N. 

and 12.5 E., and of St. Petersburg 60. o N. and 30.3 E., 
obtain the results of Ex. 3. Ans. (a) 20 ; (b) 1440 miles. 
(r) At Rome, 27.4 E. of N. ; at St. Petersburg, 43 W. of S. 



CHAPTER VIII 



FUNDAMENTAL RULES OF PROJECTION 



167. Object of chapter. In this chapter we recall cer- 
tain fundamental rules and methods of projection, developed 
in Part I., and which the reader will have learnt from his 
previous study. A collection of elementary examples is 
then given to test the knowledge of the student. These 
should present no difficulty. A well-trained student may 
generally omit them and pass on at once to the next chapter. 

168. The two principal planes of projection. Since the 
form of a simple solid is often definitely shown by two pro- 
jections only, the side vertical plane and side elevation of 
the last chapter may generally be dispensed with. There 
remain the two principal planes, the horizontal and the 
vertical planes of projection. Their intersection is called 
the ground line and will be denoted by XY ox xy. These 
planes, extended both ways, 

divide the neighbouring space -E E 

into four dihedral angles. 

A model should be made as 
follows : Take two pieces of stout 
drawing-paper, each about 9" by 6". 
Cut one along AB and the other along 

CD, CD. The latter is then folded along the lines DE, DE, passed 
through the slit AB of the former, and unfolded. The planes may then 
be turned into position at right angles to one another. 

O 




194 



PRACTICAL SOLID GEOMETRY chap. 



169. Situation in the four dihedral angles. Four points 
situated respectively in the four angles are exhibited by 
projection in the lower figures. These are derived as 
indicated in the upper figures, by first projecting the plan 
and elevation, and then turning the planes into coinci- 
dence, about XY, so as always to open out the front upper 
angle. 

We may imagine either that the vertical plane has been 
turned into the horizontal plane so that its upper portion 
moves backwards, or that the horizontal plane has been 
turned into the vertical plane, its front half falling. The 
result is the same either way. 

From the lower figures we can infer the positions of the 
points. That is, we "can measure how much the points are 
above or below the horizontal plane, and how much in front 
of ox behind the vertical plane. 

The following conception is useful in the interpretation 
of drawings. 

When considering an elevation, picture the drawing as 
coinciding with the vertical plane of projection, and think 
of the ground line xy as being an edge ox profile view of the 
horizontal plane, just as if the model of the planes were 
held on a level with the eye and viewed directly from the 
front. We thus at once recall the rule 

Rule 1. A point A is situated above or below the 
horizontal plane according to whether its elevation a is above 
or below xy. And the distance of A from the plane is 
equal to the distance of the elevation a' from xy. 

When considering a plan, picture the drawing as now 
coinciding with the horizontal plane, and conceive the 
ground line to be an edge view of the vertical plane, as if 
the model of the planes were viewed directly downwards 
from above. This at once leads to the rule 

Rule 2. A point A is situated in front of or behind 
the vertical plane according to whether it's plan a is in 
front of or behind xy. And the distance of A from the 
vertical plane is equal to the distance of the plan a from xy. 



viii FUNDAMENTAL RULES OF PROJECTION 195 




s 


Y 




i^\^ 


\ 



(4) 



x 



\a! 



(L 



y 



X 



b 



W 



y 



X 



\c 



y 



X 



d 



u- 



y 



This method of reading a drawing is often conducive to 
clearness of conception, and should be cultivated by the 
student. Instead of the planes of projection being imagined 
as turned into coincidence, they are conceived as retaining 
their horizontal and vertical positions, and it is the drawing- 
paper which is supposed to be brought to coincide first 
with one plane, then with the other, accompanied by a cor- 
responding change in the direction of view. 

Or thus, if we are reading the plan and elevation of any 
solid object, and trying thus to imagine the form of the 
latter, we think of the object as retaining its upright position, 
and regard the elevation as a picture of the object when 
viewed horizontally from the front, and the plan as its 
apparent shape when viewed vertically downwards from 
above. Thus we may regard xy as an elevation of the 
horizontal plane, or a plan of the vertical plane, or as the 
line of intersection of the two planes. 

We add a third rule. 

Rule 3. The projector drawn between the plan and eleva- 
tion of a point is always perpendicular to xy. 



196 PRACTICAL SOLID GEOMETRY chap. 

170. Rules for drawing" auxiliary projections. 

Definition i. An auxiliary elevation is the projection on 
any vertical plane not parallel to the principal vertical plane. 

This is illustrated in Fig. (i), where a" is the auxiliary 
elevation of A on the auxiliary vertical plane X' Y' . 

The lower figure exhibits this by projection, and the 
perspective view indicates how the projection may be 
derived by turning the two vertical planes backwards into 
the common horizontal plane. 

Observe that na" = ma' = aA, and that aa" is perpen- 
dicular to xy . Hence the theorem : 

Theorem i. If tivo or more elevations of a point be pro- 
jected from one plan, the distances of the several elevations 
from their respective ground lines are the same in all. 

The following simple and effective 
model should be made by all students. 

Indent and fold a piece of drawing- 
paper about 10" x 8" to represent the 
principal planes of projection, and attach 
an auxiliary plane, say 6" x 4", by paper- 
fasteners through folded margins. See the 
figure. The projections of A may be 
drawn on this model. 

Definition 2. An auxiliary plan is the projection on any 
plane which is perpendicular to the vertical plane and not 
parallel to the horizontal plane. 

Fig. (2) illustrates this case. The perspective view shows 
the point B projected on the three planes, and the arrow- 
heads indicate how two planes may be turned into the 
common vertical plane so as to obtain the lower figure. 

Observe that nb 1 = ml? = b'B, and that b"b 1 is perpendicu- 
lar to x x y v which leads to 

Theorem 2. If two or more plans or a point be pro- 
jected from the same elevation, the distances of the several 
plans from their respective ground lines are the same in all. 

The previous model will serve to illustrate this case, if it be held in 
the proper position. 

The projections having been drawn on it, the model may be viewed 




VIII 



FUNDAMENTAL RULES OF PROJECTION 



197 




at right angles to the auxiliary plane, or either hinge may be unfastened 
for the purpose of rabatment. 

Theorems 1 and 2 apply to any object, conceived as an 
assemblage of points. They may be put in the form of 
rules. Thus suppose the plan and elevation on xy to have 
been drawn ; then 

Rule 1. To obtain the auxiliary elevation on a neiv 
ground litre x'y', project from the plan perpendicular to x'y, 
and for the new elevation mark off on the projectors the dis- 
tances of the points above {or beloiv) x'y equal to the distances 
of the corresponding points of the old elevation above (or below) 
xy. 

Rule 2. To obtain the auxiliary plan on a ?iew ground 
line x x y v project from the elevation perpendicular to x x y v and 
for the new plan mark off on the projectors the distances of 
the points in front of (or behind) x x y v equal to the distances of 
the corresponding points of the old plan in front of (or behind) 
xy. 



198 



PRACTICAL SOLID GEOMETRY 



CHAP. 



X 





171. Sectional projections. An object is often assumed 

to be cut in two by a section plane, 
in order the better to show its 
internal form. A projection of 
either portion on the cutting 
plane, or one parallel to it, is 
called a sectional projection, the 
severed parts being indicated by 
section lines. 

The figure shows the plan 
and a sectional elevation of a 
regular tetrahedron. 

Note. The height of the solid is 
found by rabatting the sloping edge AD 
about its plan. 

172. Developments of surfaces. 

The surfaces of all polyhedra, and also certain curved 

surfaces, including the cone and cylinder, 

can be developed; that is, unfolded into 

one plane without any wrinkling or 

stretching. 

A workman who makes objects of 
sheet metal, or a student who makes 
paper models of geometrical solids, first 
cuts out the shape of the development. 

The illustration shows a develop- 
ment of a regular octahedron, with a margin left for 
up the edges of a paper model of the solid. 

173. Figured projections. The 

diagram shows the projection of an 
irregular tetrahedron, in which the 
numbers affixed to the letters at the 
corners indicate to a given unit or 
scale the distances of the points from 
the plane of projection. In this 
way one projection is sufficient to 
define a solid form. 




gluing 




Unit =0-1" 



viii FUNDAMENTAL RULES OF PROJECTION 199 

174. Miscellaneous Examples. 

1. Explain in your own words the meaning of the terms projector, 

projection, plan, elevation, section, planes of projection. 

2. Represent in plan and elevation four points A, B, C, and D, 

situated respectively in the four dihedral angles, each point 
being iV distant from the horizontal plane, and 2^" from the 
vertical plane. 

3. Draw the plan and elevation of a point A which is in the hori- 

zontal plane, and 1" behind the vertical plane. Also of a point 
B, in the vertical plane, and 2" below the horizontal plane. 

4. A point 1.72" below xy is both plan and elevation of a points ; 

in which of the four dihedral angles is A situated ? Find the 
true distance of A from the ground line. Ans. 2.43". 

5. Draw the plan and elevation of a point 1" below the horizontal 

plane, and 2" distant from xy. How many solutions are there ? 
Ans. Two. 

6. Determine the plan and elevation of a cube of 2^" edge : 

(a) when one face rests on the horizontal plane, and an edge 
makes an angle of 25 with xy ; (b) when an edge rests on the 
horizontal plane perpendicular to xy, and a face is inclined at 
an angle of 65 to the horizontal" plane. 

7. A cube of 2" edge is pierced by holes 1" square through all its 

faces, so as to form a framed or skeleton cube. Draw the 
plan and elevation when in the positions of Ex. 6. 

8. Draw the plan and elevation of a square pyramid, side of base 

2", length of axis 2 -J" : (a) when the base rests on the ground 
with one side making an angle 30 with xy ; (b) when a 
triangular face rests on the ground, the axis of the solid being 
parallel to the vertical plane. 

9. A point A is I-J-" above the ground, and 2" in front of the vertical 

plane. Determine the auxiliary elevation of A on a vertical 
plane which makes an angle of 50 with the vertical plane of 
projection. 

10. ABC is a triangle, the heights of A, B, and C above the 

ground being 1", 2", if", while the corresponding distances 
from the vertical plane are 1", 2J", f". In plan ab measures 
2" and be if". Draw an elevation of the triangle on a 
vertical plane which makes 50 with xy. 

11. A point A is 1" above the ground and 2" in front of the 

vertical plane. Draw the auxiliary plan of A on a plane at 
right angles to the vertical plane of projection and inclined 
at 6o to the horizontal plane. 

12. ABCD is a square. The three corners A, B, C are 2", lV, 

and l" above the ground, their distances in front of the 
vertical plane being ih", 2", I J". In plan ab is 2" and be 



200 PRACTICAL SOLID GEOMETRY ciur. 

i" long. Draw the plan and elevation of the square, and an 
auxiliary plan on a plane at right angles to the vertical plane, 
and parallel to AD. 

13. A regular tetrahedron rests with one edge (2" long) on the 
ground, whilst a face containing that edge is inclined at 40 . 
Show in plan the section of it made by a horizontal plane 1" 
high. 

14. A prism, 3" long, the ends of which are equilateral triangles 
of 1^" side, rests with a rectangular face on the ground. Draw 
the plan and a sectional elevation on a vertical plane which 
bisects the axis of the prism at an angle of 45. Draw a 
development of the portion of the surface of the solid situated 
on one side of the cutting plane. 

15. A cube, 2" edge, rests with one edge on the ground and at right 
angles to the vertical plane of projection. A face containing 
this edge is inclined at 25 ; draw the plan and elevation of 
the cube. Draw also an auxiliary elevation on a vertical plane 
which is parallel to a diagonal of the solid. 

16. Draw a sectional elevation on a vertical plane which bisects that 
edge of the cube in Ex. 15 which is on the ground, the cutting 
plane making 70 with the principal vertical plane. 

17. A hollow sphere, 2^" external diameter and i|" internal 
diameter, has a portion cut away by a plane distant J" from the 
centre. Draw a sectional projection of the larger remaining por- 
tion, on a plane which divides it into two exactly equal parts. 

18. A cube 2" edge is pierced centrally by holes |" square through 
all its faces. It rests with one face on the ground. Draw a 
sectional elevation on a vertical plane which contains a vertical 
edge and bisects a horizontal edge of the cube. 

19. A hexagonal pyramid, edge of base 1", height 3", rests with the 
base on the ground, one side of the base making io with xy. 
Draw the plan and elevation, and an auxiliary plan on a plane 
perpendicular to the vertical plane and parallel to a long edge 
of the pyramid. 

20. Suppose the pyramid in Ex. 19 to rest with a side of the base 
on the ground and perpendicular to xy, the base being inclined 
at 25 to the ground. Draw the plan and elevation, and a 
sectional plan on a plane which passes through the centre of 
the base, and is parallel to a long edge of the pyramid. 

21. Draw a line ab 2" long, and attach indices of 18 and 25 to a 
and b respectively, thus obtaining the figured plan of a line 
AB. Find the true length of AB by drawing an elevation 
on a vertical plane taken parallel to AB. Unit = o. 1". 

If ABC be an equilateral triangle having its plane vertical, 
determine the indexed plan of C, 



vin FUNDAMENTAL RULES OF PROJECTION 201 

22. Copy the figure of Art. 173 three times the size, keeping the 
indices the same. Draw an elevation of the pyramid on a 
vertical plane parallel to AC. Draw also a sectional elevation 
of the pyramid on a vertical plane through A and the middle 
point of CD. Index the plan of this section. 

23. Draw a quadrilateral, abed, making ab 2J", be 2", the angle 
abc 120 , cd 4", ad 3^"; choose v within abed, and such 
that av is if", and vd 2^". Attach indices of 12, 8, 18, 24, 
and 35 to a, b, e, d, and v. Join v to a, b, e, d; the result is 
the figured plan of a pyramid with vertex r 'and base ABCD. 
Determine (a) the plan of a section by a horizontal plane, at a 
level 19 ; (/') a sectional elevation on a vertical plane, bisecting 
AVxdA CV. Unit 0.1". 

24. Draw the plan of a cube, 2" edge, when a diagonal of the 
solid is (a) vertical, (/') horizontal. 

25. A building brick has a line joining two opposite corners 
of the solid vertical. Draw a sectional plan on a hori- 
zontal plane passing through the centre of the brick. Size 
of brick 9" x 4" x 3". Scale . 

26. An instrument box with the lid open at an angle of 120 rests 
on the ground ; draw its plan and a sectional elevation on a 
plane parallel to one end. Draw also an elevation on a vertical 
plane which makes 45 with the plane of one end. 

Dimensions of box outside length 6", breadth 4V', 
depth of box 1^", depth of lid f". Thickness of wood at 
sides and ends |-", at top and bottom j". Scale ^. 

27. Draw a development and make a paper model of each of the 
following six solids (a) a cube, (b) a regular tetrahedron, and 
(e) a regular octahedron, each of 2" edge ; (d) a building brick 
9" x 4V x 3", scale J ; (e) a prism 3" long, ends equilateral 
triangles, 2" side ; (/) a pyramid 3" high, base a square of 2" 
side. 

28. Draw plan and elevation of a regular tetrahedron, 2" edge, 
when resting with one edge on the ground and at right angles 
to the vertical plane, a face containing that edge being 
vertical. 

29. A regular octahedron, i^" edge, rests with a face on the 
ground. Draw its plan and elevation, when one edge of that 
face is parallel to the vertical plane. Draw a sectional eleva- 
tion on a vertical plane which bisects any two adjacent hori- 
zontal edges. 

30- A building brick 9" x 4^" x 3" is cut into two unequal portions 
by a plane which contains three corners. Draw the plans of 
the two parts when resting on their section faces. Index the 
plans of the corners of the solid. 



202 PRACTICAL SOLID GEOMETRY chap. 



*31. A solid is formed of two equal square prisms side of base 
lh", height 3V the axes of which bisect each other at right 
angles. The elevation is shown but is not drawn to scale. 
Draw this elevation the proper size, and deduce the plan, and 
a second elevation on a new ground line, making 50 with 
xy. (1S78) 

*32. The figure is the elevation of a truncated hexagonal pyramid. 
Draw the plan of this solid inverted, so that the plane of 
truncation is on the horizontal plane of projection. (1889) 

* 33. The position and dimensions of two equal and equally inclined 
rectangular prisms are indicated by their given projections. 
Draw the sectional elevation on AB. (1 877) 

*34. The figure represents an end elevation of an open trunk. The 
length is twice the breadth. Draw a front elevation on a 
plane parallel to one of the diagonals of the bottom. The 
thickness of wood may be neglected. (1884) 

*35. A spherical segment rests on the top of a truncated hexagonal 
pyramid. The plan and elevation are partially given. Draw 
the figure four times the given size, and make a sectional 
elevation of the solid on AB. (1888) 

"36- The plan and elevation of a desk are given. Draw a front 
elevation on a line parallel to x^y v (1886) 

*37. Make a sectional elevation on AB of the desk. (1886) 

38. Draw the plan of a square prism, height 2", side of base 
\\", a diagonal being vertical. ( 1891 ) 

39. ABC is the base of a pyramid, and Fits vertex. AB = 2,", 
AC=z", BC=2, AF=CV=3", BV=zV. Draw th 



e 



plan of the pyramid when A is ih", B2h", and C 1" above the 
horizontal plane. (1896) 



viii FUNDAMENTAL RULES OF PROJECTION 203 




CHAPTER IX 

THE STRAIGHT LINE AND THE PERPENDICULAR PLANE 

175. The possible positions of a line. The various 
positions which a line, situated in the first dihedral 
angle, may have, relatively to the planes of projection, are 
shown lettered (a) to (n). 

The remaining lines (o) to (/) are wholly or partially in 
one or more of the other three angular spaces. 

The lines (a) and (b) are respectively perpendicular -to 
the horizontal and vertical planes; (c) is parallel to the 
vertical plane and inclined to the horizontal plane ; (d) 
is parallel to horizontal plane and inclined to the vertical 
plane ; and (e) is parallel to both planes. 

The line (/) is in the vertical, plane ; (g) is in the 
horizontal plane ; and {h) is in both planes. 

The lines (k), (/), (m), (n) are inclined to both planes, 
(/) having one end in xy, (m) being perpendicular to xy, 
and (n) having its two ends respectively in the planes of 
projection. 

In studying the position of the lines, let the student 
take the model described in Art. 168. 

Pieces of wire passed through the paper, or held in the 
hand, may be used to represent the lines, and the wires 
should be placed in positions so as to agree with the pro- 
jections given. 

When the model is held with the horizontal plane on 



CHAP. IX THE STRAIGHT LINE AND PLANK 



205 




(a)(b) CO (d) (e) (f) (g) Qv) (k) 




\y/b <4> 




bb' 



CC 



< F 

'& a b 



ou 



aa. 



(l)(m){rv) (0) (p) 



(?) (T) (s) (t) 



a level with the eye, and viewed from the front, the lines 
should appear respectively as the elevations a'b' ; when 
looked down upon vertically, the appearance should be 
that of the plans ab. 



Examples. 1. Show in plan and elevation : (a) a point 2" from 
the ground line and ij" from the vertical plane of projection ; 
(b) a line parallel to and ij" from the vertical plane of pro- 
jection, and inclined at 40 to the horizontal plane. 

2. Draw the projections of any four lines, which are situated wholly 

in the four dihedral angles, one in each. 

3. Show by their projections any two equal parallel lines not 

parallel to either plane of projection. 

4. How can you tell from the projections of two lines whether the 

lines meet or not ? Draw the projections of any two lines which 
intersect, and of two lines which do not intersect. 

5. A line 2" long is parallel to xy, and distant 2" therefrom ; is 

this sufficient data to fix the position of the line ? Draw the 
projections of the line, if it be 1" in front of the vertical plane. 



206 PRACTICAL SOLID GEOMETRY chap. 

176. Problem. Having given the plan ab and eleva- 
tion ab' of a line not parallel to either plane of projection, 
to determine the length of the line, and its inclinations, 
a and /?, to the horizontal and vertical planes respect- 
ively. 

There are three useful methods of solving this problem. 

First Method. By auxiliary projections, Fig. (i). Take 
x'y parallel to ab, and draw the auxiliary elevation a"b". 
Also take x l y 1 parallel to ab', and draw the auxiliary plan 

Then both a x b x and a"b" give the length of AB. The 
angle marked a is the inclination of AB to the ground ; and 
the angle marked /3 the inclination to the vertical plane. 

Second Method. By rabatments, Fig. (2). Draw aA , 
bB perpendicular to ab, and equal to ma, nb' respec- 
tively. Draw a A v b ' B x perpendicular to ab' , and equal 
to ma, nb respectively. Then A B is the rabatment of 
AB about ab into the horizontal plane ; and A X B X is the 
rabatment of AB about a'b' into the vertical plane. 

The length of the line AB is either A Q B Q or A X B X ; and 
the inclinations are the angles marked a and f3. 

A model should be made, by cutting out the shapes of the two 
quadrilaterals in paper, leaving a margin along the edges ab and a'b' , 
for attachment to the planes of projection. 

Third Method. By turnings about projectors, Fig. (3).- 
With centre b, radius ba, describe the arc aa { , to meet 
ba x drawn parallel to xy. Project from , to meet the hori- 
zontal through a' in /. Then a'b' AB, and b'a l 'a' = a. 

With centre a', radius a'b', describe the arc b'b } ', to meet 
the horizontal through a' in b x . Project from b( to meet 
the line through b parallel to xy, in b y Then ab^ = AB, 
and ab 1 b = fi. 

The first construction represents AB turned about Bb, so that AB 
becomes parallel to the vertical plane. In the second construction AB 
is turned about Aa , until AB is horizontal. 

The lower figure shows three additional examples of this 
problem, all solved by the method of rabatments. 



IX 



THE STRAIGHT LINE AND PLANE 



207 




Examples. 1. The plan and elevation of a line are 2" and 3" 
long. The projectors of its extremities are l" apart. Find 
the true length and inclinations of the line. Ans. 3.46", 

54-7, 6o. 

2. One end of a line is .75" below the horizontal plane ; the other end 

is 1.5" above it. What is the true length and inclination of the 
line if its plan measures 2" ? Ans. 3", 48 . 

3. The elevation of a line 3" long measures 1.5" ; at what angle is 

the line inclined to the vertical plane ? Ans. 6o. 



2oS PRACTICAL SOLID GEOMETRY chap. 



177. Problem. Having given the projections of a line 
AB, to find the horizontal and vertical traces H and V. 

The plans and elevations of the traces must be respect- 
ively in the plan and elevation of the line (produced if 
necessary) ; also the plan of the vertical trace and the 
elevation of the horizontal trace must both be in xy. The 
construction is therefore as follows : 

Produce the elevation to meet xy in h' ; from h' draw a 
projector to meet the plan produced in //. H is the hori- 
zontal trace required. 

Produce the plan to meet xy in v, draw vv to meet the 
elevation a'b' produced in v'. V is the vertical trace of the 
line. 

In the right-hand figure the horizontal trace is behind the 
vertical plane. The perspective diagram underneath illus- 
trates this case. 

The construction given in this problem fails when the 
line is perpendicular to xy. In this case, find the rabat- 
ments of the line, and, if necessary, produce these to meet 
the projections produced, as shown in Fig. 6, page 207. 

When a line is parallel to a plane, its trace on that plane 
may be considered at an infinite distance away in the 
direction of the line. 

178. Problem. Having given the indexed plan of a 
triangle, to find the true shape. Unit = 0.05". 

Let abe, with the indices, be the given plan. 

Determine the rabatments of the three sides, as in the 
second method of Prob. 1 76, and draw the triangle A B Q C 
with its sides respectively equal to these rabatments. 

A B C is the true shape of the triangle, and may be 
regarded as being the rabatment of the triangle into the 
horizontal plane first, by turning the quadrilateral BCcb into 
the ground about be, then by further turning the triangle 
into the ground about -E> C . 

Examples. 1. The plan of a line is 2" long and its elevation 3". 
The projectors of its extremities are 1" apart. If the lower 



IX 



THE STRAIGHT LINE AND PLANE 



209 




end of the line be 1" in front of the vertical plane and 1^" 
above the ground, determine the horizontal and vertical traces 
of the line. 

Regarding the lower end as fixed, how many lines are there 
satisfying the above data ? Ans. Four. 

2. Draw a triangle abc having ab=2\", be =2", ac=7,". Attach 

indices of 6, 15, 24 to a, b, and c respectively. Determine 
the true shape of the triangle ABC by finding the true lengths 
of its sides. Unit = 0.1". 

Ans.- A B = 2.66", BC=2.jg", CA = $.$". 

3. Determine d U) in ac of Ex. 2. Join d and b, then DB is a hori- 

zontal line at a level 1 5. Now draw x'y at right angles to db, and 
obtain an auxiliary elevation of the triangle on x'y' ; this should 
be a straight line, the edge elevation of the plane of the triangle. 

This is a useful construction. 



210 PRACTICAL SOLID GEOMETRY chap. 

179. Solutions depending on the right-angled triangle. 
We have seen that many problems in practical geometry 
reduce to problems on the right-angled triangle. The nine 
cases of the latter have already been given in Art. 12, but 
the importance of the recognition of these analogies seems 
to warrant a repetition of the cases specially applicable to 
the problems under immediate consideration. 

Let OPp, OQq be two right-angled triangles. Then 
if OP be the true length of a line, and a its inclination to 
the ground, Op is the length of its plan, and Pp the 
difference in the vertical heights of its ends. Any two of 
these four quantities being given, the other two may be 
found by constructing the triangle OPp. 

Similarly, if OQ be the length of a line, (3 the inclination 
to the vertical plane, Oq is the length of its elevation, and 
Qq the difference in the distances of the ends of the line 
from the vertical plane. Any two of these quantities being 
given, the remaining two can be determined. 

180. Problem. A line 2" long has one end V' high ; 
the plan of the line measures 1\", and the elevation If". 
Draw the projections of the line. 

Determine by the method of Art. 179 the difference 
in the heights of the ends of the line, and in the distances 
of the ends from the vertical plane. That is, in Fig. 179 
find Pp and Qq, having given OP= OQ 2", op=\h", 
oq = if". Then proceed as follows : 

Draw any line ab, 1 |" long, as the plan. With centre b, 
radius Qq, describe a circle. From a draw am, a tangent 
to the circle, and take xy parallel to am. Then the 
elevation on xy is the one required. 

In drawing the elevation, a is to be set off |" above 
xy, and b' a distance Pp (Fig. 179) higher than a. The 
length of the elevation will then be found to be if", 
because the difference in the distances of A and B from 
the vertical plane, being equal to bm, is equal to Qq 
(Fig. 179). 



IX 



THE STRAIGHT LINE AND PLANE 



211 





S^< ^7 \ 




180 v n 



/ 



./ 



Examples. 1. Take a point a in xy, and draw ab I j" long, perpen- 
dicular to xy. Let ab be the plan of a line AB, of which B 
is I V higher than A ; find the length and inclination of AB. 
If A be Jf " above the ground, draw the elevation of AB, and 
find the horizontal trace of the line. 

2. A point A is 2" above the ground and 1^" in front of the vertical 

plane. AB is a line 2|" long, whose plan is perpendicular to 
xy ; draw the plan and elevation of AB, when B is in the 
vertical plane. 

3. The plan of a line measures 2" and the elevation 1^". Is this 

data sufficient to fix the length of the line? If not, add any 
third condition which will fix the length. 

A is a point in the vertical plane ij" above the horizontal 
plane ; B is a point in the horizontal plane i|" from the vertical 
plane. The real distance from A to B is 3". Draw the plan 
and elevation of the line AB. 

4. A point is 2" from the vertical plane and 1.75" above the hori- 

zontal plane. Determine a point B on the ground line 3j" 
from A, and measure the inclination of AB. 

5. Two points A and B are respectively 1.5" and 1.75" from 

both planes of projection. Their plans are 2.25" apart. De- 
termine the projections of a point C in the vertical plane 2" from 
A and 2.75" from B. If C has to be 2" from A, what is the 
least possible distance of C from B? Ans. 2". 

6. Determine the projections of a point D in AB, Ex. 5, distant 

1.75" from A. 

7. A line AB is inclined at 35". The upper end A is 2" above the 

ground, and ab is 2" long. Draw the elevation of AB on x'y' 
which makes 30 with ab. 

8. A line 3" long has one end 1" high, the plan of the line measures 

2" and the elevation 2j". Draw the projections of the line. 



212 PRACTICAL SOLID GEOMETRY chap. 

181. Problem. A line AB of given length has its ends 
in the planes of projection, and the line is inclined at 
given angles a and 6 to the horizontal and vertical planes 
respectively. Draw the plan and elevation. 

Method i. By the construction of Art. 179 this problem 
could be at once converted into the case of the preceding 
problem, and thus solved. 

Method 2. Draw B x d and B x c inclined at a and /3 to xy, 
and make B x d = B x c = AB. Draw the projectors da, cc. 

Then, as in Art. 179, B x a is the length of the plan of 
the line, and B x c the length of the elevation. 

Make db' = B x c' , and from b' draw a projector to meet 
the circle, with a as centre, and aB x as radius, in b. Then 
ab, db' are the required projections. 

To explain this construction, suppose in the first instance the line to 
be in the vertical plane, so that aB v a B x are its projections. Let the 
line be then turned about ad as axis. The path of B is a circular arc 
of which B x l> is the plan, and B x b' the elevation. The rotation is 
continued until the length of the elevation is equal to B x c ; the line is 
then inclined at /3 to the vertical plane. It is also inclined a to the 
horizontal plane, since the latter inclination remains unchanged during 
the rotation. 

Limits to data. The sum a+/5 cannot be greater than 90 . 

182. Problem. Having given the projections of a 
line AB, to find the shortest distance of the line (produced 
if necessary) from xy. 

Let ab, db' be the given projections. 

Draw 0J1 and o v, an edge view of the horizontal and ver- 
tical planes of projection. 

Determine a Q , the projection of A in this view, by 
making a m = a'/, that is, equal to the height of A above 
the ground; and a Q n = a/, that is, equal to the distance of 
A in front of the vertical plane. 

Similarly determine b and join a Q > . Draw o^ Q perpen- 
dicular to a b . Then o c is the shortest distance required. 

Obtain c from c , and c from c ; then co, c'd are the 
projections of the shortest line between xy and AB. This 
line is perpendicular to both xy and AB. 



IX 



THE STRAIGHT LINE AND PLANE 



213 




Examples. 1. Draw the plan and elevation of any line 2\" long, 
which is inclined at 35 to the horizontal plane, and 25 to the 
vertical plane. 

2. A line 2^" long has one end in xy, and is inclined at 30 and 

40 respectively to the horizontal and vertical planes of projec- 
tion. Represent the line. 

3. Draw the plan and elevation of any line 3" long, which shall be 

equally inclined to the planes of projection, and not parallel 
to xy. 

4. The plan of a line 2.\" long measures 2" ; draw the elevation on 

a vertical plane inclined to the line at 20 . 

5. The plan of a line measures 1" and the elevation .8". The in- 

clination of the line to the horizontal plane is 30 ; what is the 
inclination to the vertical plane? Ans. 46. 2. 



6. Draw a triangle abc having ab = 2^", ac = \\ 



7. 
8. 
9. 



be 



Through 



draw cd 3" long, cutting ab and making an angle of 50 
with ca. Attach indices of 8, 18, 6, 15 to a, b, c, d. 
Find whether the lines AB and CD intersect. Unit = o. 1". 

If they do not, find what the index of d must be so that 
the lines shall intersect. Ans. 25.6. 

Draw the plan of a horizontal line at a level 10 to intersect 
the lines AB, CD in Ex. 6. 

A point A is Y above the ground, and 1^" behind the 
vertical plane. Find the distance of A from xy. Ans. 1. 58". 

A point A is 1" above the ground, and ii" in front of the 
vertical plane. A point B is h" above the ground, and 2|" in 
front of the vertical plane. The projectors of the points are 
1 f" apart. Find the shortest distance between AB and xy. 
Draw the projections of the shortest line between AB and xy. 



214 PRACTICAL SOLID GEOMETRY chap. 

183. The perpendicular plane. A perpendicular plane is 
one which is at right angles to one or other (or both) of the 
principal planes of projection ; a plane which is inclined to 
both being called oblique. A perpendicular plane may be 
either inclined or vertical. An inclined plane is one which 
is inclined to the horizontal plane and perpendicular to the 
vertical plane. A vertical plane is perpendicular to the 
horizontal plane and may be inclined to the vertical plane 
of projection. 

As was stated in Art. 166, a plane is conveniently repre- 
sented by its traces. We shall now have only two traces, 
the horizontal and the vertical. 

The student should make a model of a perpendicular plane as shown 
at (a). For the planes of projection take a piece of drawing- 
paper 9" square, indented and folded across the middle. For the per- 
pendicular plane, cut a rectangular piece of drawing-paper 6|" by 5j", 
and indent and fold margins |" wide on two adjacent edges for attach- 
ment by paper-fasteners. By taking out one or other pair of fasteners, 
the plane can be rabatted into either plane of projection at pleasure ; 
and by placing the model the two ways up in succession, it illustrates 
both the inclined and vertical plane, as shown at (a) and (/'). 

The characteristic feature of a perpendicular plane is 
that one or other of its traces is a profile or edge view of the 
plane. In the oblique plane neither trace is so. 

To illustrate this let the student take the model, and 
holding it in the position (a), let him view it directly from 
the front ; the whole plane will appear as a line coinciding 
with the vertical trace. The vertical trace is thus strictly 
an elevation of the plane, for it contains the elevations of all 
points and lines in the plane. The horizontal trace could 
not be called a plan of the plane, for it contains the pro- 
jection of only one line in it, the trace itself, as will be quite 
evident from the model. 

Similarly in (/>), it will be at once seen from the model 
that the horizontal trace is a true plan of the vertical plane, 
for it contains the plans of all points and lines in the plane. 
But the vertical trace is not an elevation. 



IX 



THE STRAIGHT LINE AND PLANE 



215 





v v 




V 



-v 



h 



(a) (7? j (C) 



(d) 



y 



h 



Cej 



Five perpendicular planes are represented by their traces 
in the lower figure : (a) as an inclined plane, (b) a vertical 
plane, and (c) a plane perpendicular to xy ; (d) is a hori- 
zontal plane, and (e) a plane parallel to the vertical plane 
of projection. In (a) tv is an edge elevation of the plane ; 
in {b) th is a profile plan ; and in (1), (</), and (e) the traces 
are all edge views of the planes. 

By using the model the student will see that the angles 
which the planes make with the horizontal and vertical 
planes of projection are respectively equal to the angles 
which the vertical and horizontal traces make with xy. In 
an oblique plane this is not the case. 

It will also be noticed that the apparent angle between 
the traces is not the real angle, for the latter is a right 



angle. 



216 PRACTICAL SOLID GEOMETRY chap. 

181 Problem. A square pyramid is given by its 
projections, (a) Determine the plan of the section made 
by any inclined plane VTH. (b) Draw the plan of the 
frustum with its section end on the ground, (c) Draw a 
development of the surface of the pyramid, showing the 
trace of the cutting plane on the surface. 

(a) The elevations of P, Q, P, S, the points in which 
the plane cuts the edges, are on tv, because the vertical 
trace is an edge elevation of the plane. 

The plans/, q, r, s are obtained by projection from the 
elevations. 

(b) On tv draw an auxiliary plan of the frustum, as 
shown in the figure ; this is determined according to the 
second rule of Art. 170, and is the plan when standing on 
the section end. 

The plan may also be considered as obtained by rabat- 
ment of the plane about its vertical trace, the frustum 
having been first projected on the section plane. 

(c) To obtain a development, first find the true length 
of a sloping edge of the pyramid and the true distances of 
P, Q, P, S from the vertex. In the figure v'd^ and v'q( 
are the true lengths of VD and VQ respectively, obtained as 
in Prob. 176, third method. The lengths of FP, VP, 
VS may be found in a similar manner. 

A development of the surface of the pyramid must now 
be drawn as shown, and the lengths VQ, VP, VS, and 
VP set off along the developed edges. The development 
is then completed by joining QP, PS, SP, PQ, these lines 
forming the developed trace. 

Examples. 1. A cube, 2" edge, has its base on the horizontal 
plane, one corner being in xy, and the diagonal of the base 
through that corner making 6o with xy. Draw the plan and 
elevation showing the section by a plane passing through the 
centre of the cube, and inclined at 30 . Find the true shape of 
the section, and draw a development of the frustum. 

2. A regular tetrahedron ABCD of 3" edge rests with ABC on the 
ground. Draw its plan and show the section by a plane through 
the centre of the solid, whose horizontal trace is am making 
45 with ab. 



IX 



THE STRAIGHT LINE AND l'LANE 



217 




-y 



b 




218 PRACTICAL SOLID GEOMETRY chap. 

185. Problem. Having given the projections of a cube, 
and the traces vth of a vertical plane, to determine the 
plan, elevation, and true shape of the section. 

The horizontal trace lit is an edge view of the plane, and 
the required plan of the section is therefore in the horizontal 
trace, and is the line pq. The elevation of the section is 
the shaded rectangle obtained by projection from pq. 

The true shape of the section is the rectangle pQ m 
obtained by the rabatment of the section plane into the 
horizontal plane, pP being made equal to p'p'. 

The true shape might be equally well obtained by a 
rabatment into the vertical plane, about the vertical trace, 
and the student should make this construction. 

Example. A square pyramid, side of base i|", axis 2^", has its 
base in the vertical plane, one corner being in xy, and the diagonal 
of the base through that corner inclined at 6o. Draw the plan 
and elevation, showing the section by a vertical plane, bisect- 
ing the axis of the solid, and making 30 with the vertical 
plane of projection. Find the true shape of the section. 

186. Problem. Having given the plan of any polygon 
or plane figure lying in a given inclined plane, to find the 
elevation, rabatment, and true shape of the figure. 

Let abc be the given plan, and vth the traces of the 
given plane. 

From the points in plan draw projectors to meet the 
vertical trace in a, b', c . Then db'c is the required 
elevation of the polygon. For, the vertical trace being an 
edge elevation of the plane, it must contain the elevation of 
the polygon. 

The true shape may be obtained by a rabatment of the 
plane about its horizontal trace. With centre / describe the 
arcs a'A ', b'B ', c'C '; these are the elevations of the 
paths of A, B, C; and A ', B \ C ' are the elevations of 
the rabatments of A, B, C. The plans of the arcs are the 
lines aA , bB , cC , drawn perpendicular to the horizontal 
axis of rotation th. These intersect the projectors from 
A ', ', Q in A v B , C . 



IX 



THE STRAIGHT LINE AND PLANE 



219 



>>.v 




% \C y 



The true shape may also be obtained by a rabatment 
into the vertical plane ; this is equivalent to an auxiliary 
plan on tv, and may be determined by applying the 
principles of Art. 170. 

This is an important problem because it occurs so 
often in other problems ; hence the student should make 
himself quite familiar with the construction. 

Examples. 1, Draw an equilateral triangle abc, 2" side, with ab 
making an angle of 45 with xy. Let this be the plan of a 
triangle ABC lying in a plane inclined at 40 . Find the in- 
clinations of the sides of the triangle. 

2. By using the rabatment of the triangle ABC in Ex. 1, determine 
(a) the plan of the bisector of the angle ACB ; (/') the plan 
of the perpendicular drawn from C to AB. 



Note. 



-Observe that if in the rabatment the angle A C Q B n be 



bisected by a line which meets the horizontal trace of the plane 
in ?', then ic is the plan of the bisector. In problems in- 
volving rabatments it is a very useful artifice to produce lines 
to meet the trace in stationary points. 



220 PRACTICAL SOLID GEOMETRY chap. 

187. Problem. To find the point of intersection of a 
given line AB and inclined plane VTH. 

The elevation of the required point of intersection is 
/', for reasons stated in previous problems. The plan p is 
found by drawing the projector from /' to meet ab. 

188. Problem. To determine a perpendicular from a 
given point A to a given inclined plane VTH. 

Draw a'm perpendicular to tv ; through ni draw the 
projector mm, to meet at m a line through a perpendicular 
to th. Then AM is the required perpendicular. 

The student should use the model to illustrate this problem. 

189. Problem. To determine the plan and elevation of 
the projection of a given point A on a given inclined 
plane VTH. 

Proceed as in the last problem. Then M is the required 
projection. 

190. Problem. Having given a line AB and an inclined 
plane VTH, to determine (a) the trace of the line on the 
plane ; (b) the projection of the line on the plane ; and 
(c) the angle between the line and plane. 

(a) The required trace of the line on the plane is the 
intersection S, and is found as in Prob. 187. 

(b) The projection Jl/JV 'of the line on the plane is de- 
termined by applying the construction of Prob. 189. 

(c) The inclination of the line to the plane is the angle 
between the line AB and its projection MN on the plane, 
and is found by a double rabatment as follows : 

Fust, MqNq, the rabatment of MN, is found as in Prob. 
186. Then A B Q JV Af Q , the rabatment of the quadrilateral 
ABMJV about M N Q , is determined by drawing M A , N B Q 
perpendicular to M Q 2V Q and equal to am, b'ri, the true 
lengths of the perpendiculars AM, BN. 

The inclination of the line to the plane is the angle 

AqSqMq. 

The student may readily make a model to illustrate this problem, 
by cutting out the shape of AqBqMqN'q in paper, with a margin at 
MqNq for attachment to the model of the inclined plane. 



IX 



THE STRAIGHT LINE AND PLANE 



221 





TTL 



188 



Oj 



k 




222 PRACTICAL SOLID GEOMETRY chap. 

191. Problem. Determine the plan and elevation of a 
line inclined at a, and lying in a plane inclined at e. 
Find the angles which the line makes with the traces of 
the plane, and the inclination of the line to the vertical 
plane. 

Draw the traces vth of the plane inclined at 9. 

Take any point A in the vertical trace of the plane ; its 
elevation a will be in tv, and its plan a in xy. 

Through A draw a line in the vertical plane inclined at 
a, with one end B on the ground, so that a'B 1 is its eleva- 
tion, and a 1 its plan. 

Let this line revolve about ad until its end B comes into 
the horizontal trace of the plane ; the line itself is then in 
the plane, because / both its ends are in it. The arc B Y b 
struck with a as centre is the plan of the path of B, and 
B x b' is the elevation of the path. 

Then ab, db' are the projections of the line in the re- 
quired position ; for the inclination of the line remains un- 
changed during the rotation. 

Let bA be the . rabatment of the line obtained as in 
Prob. 1 86, then the true angles which AB makes with the 
horizontal and vertical traces are respectively A bb' and 
bA b\ The latter angle is also /i, the inclination of AB to 
the vertical plane, since it is equal to the angle between 
the line and its elevation. 

This problem is an important one, and should be studied 
with the model, until the exact meanings of the five angles 
referred to are fully understood, and the differences between 
them are recognised. Students partially acquainted with 
the construction are apt to take wrong centres for the 
two arcs. 

Limits to the data. The angle a cannot be greater than 6, but may 
have any value from O to 6. When a = o, the line is horizontal, and 
therefore parallel to the horizontal trace of the plane. When a = 0, 
that is, when the line and plane are equally inclined, the plan of the 
line is perpendicular to the horizontal trace, and the line itself in the 
plane is also perpendicular to the horizontal trace. 



IX 



THE STRAIGHT LINE AND PLANE 



22' 



./ 




191 
192 



192. Problem. Determine the plan and elevation of a 
line inclined at 6 to the vertical plane, and lying in a 
plane inclined at e to the horizontal plane. Find a, the 
inclination of the line to the ground. 

Let vth be the traces of the plane inclined at 0. 

First draw bA Q making an angle j3 with xy, and suppose 
this to be a line in the horizontal plane. 

Now let this line be turned about bt until it comes into 
the plane VTH. That is, with centre / describe the arc 
A a\ and draw the projector a a. 

Then ab, a'b' are the projections of the line in the 
required position. 

The inclination a is obtained by turning AB into the 
vertical plane, about a a, as in Prob. 176, third method. 

The construction of this problem corresponds to that of 
the preceding problem worked backwards, and the two 
problems should be studied together with the model. 

Example. A vertical plane makes 45 with xy. Draw its traces 
and show a line lying in it and inclined at 45. Find the 
angle which this line makes with the vertical plane of projection. 



224 PRACTICAL SOLID GEOMETRY chap. 

193. Plane represented by a scale of slope. It lias 
already been explained that by a system of indexed plans, 
points and lines may be represented by one projection 
only. It has now to be shown how in the same system a 
plane is represented. 

Draw a line through any two points, P, Q, on a piece of cardboard, 
and let one edge A B of the cardboard be perpendicular to PQ. Take 
the cardboard as the model of a plane, and hold it in an inclined 
position, with AB resting on any flat surface representing the horizontal 
plane. 

The student will then observe that PQ is inclined at 
the same angle as the plane, and that the plan of PQ is 
perpendicular to the horizontal trace AB. Also that any 
line parallel to PQ, and lying in the plane, has the same 
inclination as the plane. 

Lines in the plane such as PQ, perpendicular to the 
horizontal trace, are called lines of slope, and it is evident 
that if the figured plans of P and Q, or of any two points 
on any line of slope, be given, the position of the plane 
relatively to the horizontal plane is completely defined. 

The figured plan of a line of slope is called a scale 
of slope, and is conventionally drawn as two lines, one 
thicker than the other, as is usual in drawing scales. The 
double line serves to distinguish the representation of a 
plane from that of a line. 

Fig. 193 shows the representation of a plane by a scale 
of slope, the unit for the indices being 0.1". 

To represent this plane in the manner previously 
explained, see the next problem. 

194. Problem. Having given a line by its indexed 
plan, a 12 b 5 . 5 , and a plane by its scale of slope, 5,15, to deter- 
mine the indexed plan of the point of intersection of the 
line and plane. 

Draw any xy parallel to the scale of slope, and find 
db', the elevation of AB ; find also 5', 15', the eleva- 
tions of the two given points on the scale of slope. 



IX 



THE STRAIGHT LINE AND PLANE 



225 




193 




Then vt, drawn through 5', 15', is the vertical trace of 
the given plane as regards xy ; and p' is the elevation of 
the intersection of the line and plane, as in Prob. 187. 

Project from /' to determine the plan p. To find 
the index of p, measure the height of/' above xy, or read 
the position of q on the scale of slope. 

Example. Draw a line ab 1^" long ; draw ad and be each perpen- 
dicular to ab and I" and i"in length. Attach indices of 10, 30, 
25, and 15 to a, b, c, J. Regard ab as the scale of slope of a 
plane and cd the figured plan of a line. Determine the indexed 
plan of/', the intersection of the line and plane. Unit 0.1". 

195. Problem. Having given the indexed plan of a 
triangular pyramid, and the scale of slope of a plane, to 
determine the indexed plan of the section of the solid. 

(No figure.) 

Draw an elevation of the plane and pyramid on an xy 
taken parallel to the scale of slope, and proceed as in 
Prob. 184. Measure the heights of the points in the 
section, and index the corresponding points in plan. 

Q 



226 PRACTICAL SOLID GEOMETRY chap. 



196. Miscellaneous Examples. 

1. Draw the traces of a perpendicular plane, inclined to the right at 

40 to the ground ; draw ab 1" long anywhere to the right of 
the horizontal trace. Regard ab as the plan of a line AB lying 
in the plane, and determine the plan of a square ABCD also 
lying in the plane. 

Hint. First find the rahatment A B C D ; produce C B to 
meet the horizontal trace in i ; join ib and produce it to meet in c 
a line from C parallel to xy ; then complete the parallelogram 
abed. Observe that by working with i we do not require to 
use the elevation in order to determine c from 6? . 

2. Represent two planes inclined, one at 40 and the other at 70 , 

and such that their intersection is I J" from the ground. Obtain 
the projections of a horizontal line, i" above the ground and 
2V' long ; the ends of which are in the two planes. Determine 
the point of intersection of this line with the inclined plane 
which bisects the acute angle between the first pair of planes. 

3. Determine the traces of a plane inclined to the right at 15 , 

and perpendicular to the vertical plane. A point A is \\" in 
front of the vertical plane, and 2" above the horizontal trace, 
but 1" to the right of it. Draw the projections of the per- 
pendicular let fall from A on to the plane. 

4. Represent a vertical plane making an angle of 40 with xy. 

In this plane place a line 2j" long, which shall be inclined to 
the vertical plane of projection at an angle of 30 . 

5. Represent a vertical plane which makes an angle of 40 with.rj', and 

draw the projections of a square lying in this plane, one diagonal 
being inclined at 70 , the lower end of which is on the ground. 

6. An inclined plane makes an angle of 40 with the ground ; 

draw the projections of a line which bisects the angle between 
the traces. 

7. Draw the plans of two lines AB, AC, which lie in a plane 

inclined at 50 , the lines being inclined at 30 and 45 
respectively. Obtain the projections of the bisector of the 
angle BAC. 

8. Draw the traces of a plane inclined at 50 , and in this plane 

place a line 2" long inclined at 40 . Find the inclination 
of the line to the vertical plane. 

9. The horizontal trace of a vertical plane makes 42 with the 

ground line. Determine the elevation of a line lying in this 
plane, inclined at 30, and passing through the point where the 
given plane cuts the ground line. 
10. Draw the traces of a vertical plane which makes an angle of 6o 
with the vertical plane of projection. Represent a line lying 
in this plane, and inclined at 70 to the vertical trace. 



ix THE STRAIGHT LINE AND PLANE 227 

11. A person on the top of a tower 60 feet high, which rises from 
a horizontal plane, observes the angles of depression of two 
objects A and B on the plane to be 20" and 30, the directions 
of A and B from the tower being west and south respectively. 
Find (a) the distances of A and B from the tower ; (/>) the 
distance apart of A and B ; and (c) the direction of B and A. 

12. Represent a plane inclined at 50 , and in it place two lines, 
one horizontal and the other inclined at 30 . Find the true 
angle between these lines. 

13. The face of a hill is inclined at 30 , the lines of slope being 
due east ; what is the inclination of a path on the hillside which 
goes in a north-easterly direction. 

14. A plane inclined at 40 contains a point A, distant 1" and ih" 
respectively from the horizontal and vertical planes of projec- 
tion. Determine the rabatments of the point about the traces 
of the plane. 

15. A line is inclined at 35 and 50 to the horizontal and vertical 
planes respectively, its traces are 4" apart. Determine its 
projections. 

16. A line ab 2" long, making 30 with xy, is the plan of a 
horizontal line 1 j" above the ground. Find its elevation, and 
determine the plan and elevation of an isosceles triangle, having 
the given line for base, and its vertex in xy. 

17. Two points are respectively 1.5" and 0.75" from both planes 
of projection. Their plans are 2.25" apart. Determine a 
point on the ground line equidistant from the points. 

18. A rectangle, sides 2" and 3", revolves upon one diagonal as a 
fixed horizontal line until the plan of a right angle opposite 
becomes 120 . What is then the inclination of the plane of 
the figure ? 

19. Two lines inclined to the horizontal plane at angles of 25 and 
45 respectively are drawn from a point situated 2 " from both 
planes of projection. The plans of these lines make 110 
with each other. Determine the real angle between the 
lines. 

20. Given a point 1.25" from the horizontal plane, and 1" from the 
vertical plane ; obtain the projections of any line 4-I" long 
passing through the given point and terminated by the planes 
of projection. 

21. A line is inclined at 30 and 40 to the planes of projection ; 
what is its inclination to a plane which is perpendicular to xy ? 

22. Draw the plan of a regular octahedron of 1 h" edge when resting 
with one face on the ground. Determine the plan and true 
shape of the section made by a plane which is inclined at 
45 and contains one diagonal of the solid. 



228 PRACTICAL SOLID GEOMETRY chap. 



""23. Determine the inclination of the line AB (Fig. a) to each plane 
of projection. 

K "24. Determine the horizontal traces, //"and K, of any pair of lines 
which pass through P (Fig. a), and intersect AB in, say, C and 
D. Show in the same figure the true shapes of the triangle 
PHK and the quadrilateral CD// A'. 

*25. From a given point/,/' (Fig. a), draw a perpendicular on the 
given line ab, a'b'. ( 1 877) 

Hint. Proceed as in Ex. 24, and having obtained the true 
shapes PqH^A'q and C D H A' , draw a line from P perpen- 
dicular to CqDq, meeting A/ A" in Z . The projections of the 
required perpendicular may be at once obtained from Z Q . 

*26. A line parallel to the vertical plane is given by its projections 
ab, a'b' (Fig. b). Draw the traces of a plane containing this line 
and perpendicular to the vertical plane. In this plane draw 
lines passing through A and B and making 45 with the line 
AB. Draw also an elevation of the three lines on a vertical 
plane perpendicular to the ground line. (1885) 

*27. a'b' (Fig. c) is the elevation of a line 2!" long. Its centre C 
is 2" from xy. Determine (1) the plan of C, (2) the difference 
of the distances of A and B from the vertical plane of 
projection. 

*28. a'b' (Fig. c) is the elevation of a line 2i" long. Its centre 
point is 2" from xy. Draw its plan. (1884) 

*29. (Fig- d) is the lowest corner of a cube ; ab is the plan of 
one edge, AB (real length = 20 units), of a face which lies in 
a plane, of which ht is the horizontal trace. Draw the plan 
of the cube. Also the plan of its section by a plane parallel 
to the horizontal plane of projection, and at a height of 12 
units above it. Unit = o. 1". (1896) 

*30. Determine the height of P, a point on CD, whose figured plan 
is given (Fig. <?). Unit = 0.05". 

*31. In Fig. (e) p is the centre of a sphere 2\" radius. Determine 
a sectional elevation of this sphere on a vertical plane whose 
horizontal trace is ab. Unit 1=0.05". 

""32. Two lines are given by their figured plans, Fig. (e). From 

the point P on one of them draw a line 2 -J" long, terminating 

on the other. Unit = 0.05". (1886) 

Hint. Proceed as in Ex. 31, obtaining an elevation of AB 

on the vertical plane whose horizontal trace is ab. 

*33. Find the length of the line CD in Fig. (/), and determine the 
length of the diagonal of a cube of which CD is one edge. 
Unit = 0.1". 

*34. Determine whether the triangle CDE, Fig. (/), is right-angled 
at D. 



IX 



THE STRAIGHT LINE AND PLANE 



229 




CHAPTER X 

THE OBLIQUE PLANE 

197. Introduction. As defined in Art. 1S3, an oblique 
plane is one which is inclined to both planes of projection. 
The traces are both inclined to xy, or both parallel to it, or 
both coincide with it ; and the true angle between the 
traces is not a right angle. 

The figure shows four cases of an oblique plane. In 
(a) the real angle between the traces is acute ; in (b) it is 
obtuse ; in (c) the traces are parallel to one another ; and in 
(d) the plane contains the ground line, and cannot be repre- 
sented by its two traces in the ordinary way, since these 
both coincide with xy. It requires to be shown by a side 
elevation, on the plane perpendicular to xy, as indicated 
in the figure. 

Let the student illustrate these cases by four models, made in the 
manner explained in Art. 183, and shown in the figure. The planes 
are attached to the planes of projection by paper-fasteners passing 
through folded margins ; either margin can be unfastened to allow of 
the plane being rabatted about the other trace. 

The student should note that the intersection of the 
horizontal and vertical traces of a plane is always on the 
ground line xy. When one trace is parallel to xy, so is the 
other, and the intersection of the three lines is at an infinite 
distance away in the direction of the ground line. 



CHAP. X 



THE OBLIQUE PLANE 



2 3i 




(a) &> h (b) 



(C) 






A characteristic feature which distinguishes an oblique 
plane from one which is perpendicular to one or both of the 
planes of projection, is that neither trace is an edge or pro- 
file view of the plane. On this account the problems are 
generally more difficult than those of Chap. IX. We shall 
begin with some cases which may be made to depend on 
the constructions of Chap. IX. 



232 PRACTICAL SOLID GEOMETRY chap. 

198. Problem. To convert a given oblique plane into 
an inclined plane, by means of an auxiliary elevation. 

In the diagram VTH \s, the given oblique plane. 

Take Mn, a new vertical plane of projection, at right 
angles to the given plane. Then X'Y,, the new ground 
line, is perpendicular to TH, the horizontal trace. 

With reference to this new plane of projection the given 
plane is an inclined plane (Art. 183). 

Let ON be the intersection of Mn and J'TH Then 
ON is the new vertical trace, and is also an edge elevation 
of the plane VTH. 

The above process has now to be represented by ortho- 
graphic projection. 

Let vth be the given traces of the plane. 

Take a new ground line x'y perpendicular to th, inter- 
secting th in and xy in n. Draw nn, nn" perpendicular 
to xy, x'y, and make nn nn. Thus n" is the auxiliary 
elevation on x'y' of the point N. And on" is the new 
vertical trace required. 

The student should make a special model to illustrate this problem, 
by cutting a model of the oblique plane along a line NO perpendi- 
cular to TH, and inserting the new plane of projection Mn. 

On holding the model so as to look along HT, it is 
seen that X' Y is the new elevation of the ground, and NO, 
the new vertical trace, is the edge elevation of the plane. 

The lines nn, nn" are the positions taken by nN, the 
line of intersection of the two vertical planes of projection, 
when these planes are turned back about their respective 
ground lines into the horizontal plane. 

It should be observed that the angle v'on is the true 
inclination, 6, of the plane to the ground. 

Figs. (l>) and (c) show two other examples of this 
problem. In Fig. (l>) the plane vth is the same as in Fig. 
(a), but x'y' is drawn in a different position, to avoid the 
overlapping of the two elevations. Fig. (r) shows what 
form the construction takes when the real angle between 
the traces is obtuse. 



X 



THE OBLIQUE PLANE 



233 




By means of the construction just given, many problems 
on the oblique plane may be at once converted into cor- 
responding problems on the inclined plane, and thereby 
simplified. This method of attacking such examples is a 
most valuable one, on account of its wide applicability, 
and should become quite familiar to the student. The 
six problems which immediately follow are worked in this 
manner. 

Example. Convert the following oblique planes into inclined 
planes by means of auxiliary elevations, and in each case 
measure the inclination to the horizontal plane : 

(a) The horizontal and vertical traces ///, tv make angles of 

40 and 55 with xy. 
(/>) The traces make 120 and 35 with xy. 
(<) The traces are in one straight line making 45 with xy. 
Am. (a) 65.8 ; (6) 39^ : (<r) 54.7". 



234 PRACTICAL SOLID GEOMETRY CHAP. 

199. Problem. To determine the point of intersection 
of a given straight line AB, and a given plane VTH. 

Take x'y' perpendicular to th, and on x'y draw elevations 
of the plane and line as shown. 

The problem is thus reduced to that of Prob. 187, 
c" being the auxiliary elevation of the point of intersection 
of the line and plane. 

The plan c is obtained by projecting from c" : and the 
elevation c by projecting from r. 

200. Problem. To determine the plan and elevation 
of the section of a given prism by a given oblique plane 
VTH. Also to draw a sectional plan of the solid. 

By drawing an auxiliary elevation, this problem is con- 
verted into that of finding the section of a solid by an 
inclined plane, and is then worked as In Prob. 1S4. 

Take x'y' perpendicular to tA, and on x'y' draw the 
auxiliary elevations of the plane and prism as shown. 

Then ov' is an edge elevation of the plane, and a ", />", c" 
are the auxiliary elevations of the points where the plane 
cuts the three vertical edges of the prism. 

The heights of a', />', c above xy are now made equal to 
the heights of </', />", c' above x'y', thus determining the 
elevation of the section. The plan of the section is abc, 
coinciding with the plan of the prism. 

The required sectional plan is a projection of one portion 
of the solid on the cutting plane, or on a plane parallel to it. 

Take ov as a second auxiliary ground line, and from 
the auxiliary elevation project the plan a^/ v making the 
distances of a v /> v c x from ov the same as the distances of 
a, b, c from x'y, according to the rules of Art. 170. 

The sectional plan is completed by drawing the plans 
of the other lines of the frustum, obtained in a similar 
manner. 

Compare this plan with the plan of the frustum of the 
pyramid in Prob. 184. The two are obtained by the appli- 
cation of the same principle. 



X 



THE OBLIQUE PLANE 



235 




/A 



Examples. 1. Draw the traces of the three planes of the example 
on page 233. Draw the projections of a line parallel to xy 
and ii" from each of the planes of projection. Determine the 
plan and elevation of the intersection of this line with each of 
the three planes. 

2. Draw the plan and elevation of a square pyramid, side of 
base 2", height 3", the base being on the ground with its 
centre if from xy, and one edge making 35 with xy. 
Select a point on xy, 3-J" to the left of the plan of the centre 
of the base. Through this point draw horizontal and vertical 
traces making 45 and 38 respectively with xy. Determine 
the plan and elevation of the section of the pyramid made by 
this plane. Draw also a sectional plan of the pyramid. 



236 PRACTICAL SOLID GEOMETRY chap. 



201. Problem. To determine the perpendicular from 
a given point A to a given oblique plane VTH. 

Convert the problem into that of Prob. 188 by 
drawing ov' and a", the auxiliary elevations of the plane 
and point on x'y', taken perpendicular to th. 

Through a" draw a" m" perpendicular to ov ; then a" m" 
is the auxiliary elevation of the perpendicular. 

Through m" draw the projector m"m to intersect in m 
the line through a perpendicular to th. 

The required plan, am, is thus determined ; and the 
elevation on xy is at once found by projection from the 
plan, since the height of M is known. 

It may be proved by pure solid geometry that 

Theorem. If a line and plane be perpendicular to each 
other, the plan and elevation of the line are respectively per- 
pendicular to the horizontal and vertical traces of the plane. 

The student may satisfy himself of the truth of this 
proposition by using a model. The accuracy of the above 
solution should be tested by ascertaining whether am is 
perpendicular to tv, as ought to be the case. 

202. Problem. To determine a plane which shall bisect 
a given straight line AB at right angles. 

Take x'y parallel to the given plan ab, and draw the 
auxiliary elevation d'b". Draw p"o bisecting a"b" at right 
angles, and meeting x'y in o. Then p"o is an edge eleva- 
tion on x'y, of the required plane. 

The horizontal trace of the plane is ot, drawn through 
o perpendicular to ab ; and the vertical trace is tv, drawn 
through / perpendicular to ab'. 

203. Problem. To determine the true distance be- 
tween two given parallel oblique planes. (No figure.) 

If the planes are parallel, their horizontal and vertical 
traces are also respectively parallel (Theorem 13, Appendix). 

The required distance between the planes is that be- 
tween their edge elevations, obtained as in Prob. 198. 



THE OBLIQUE PLANE 



237 




Examples. 1. Represent an oblique plane, and a line perpendi- 
cular to it. Show also an oblique plane and a line lying in it. 

2. Draw the traces of the three planes of the example on page 233, 

and in each case draw the projections of a point A, situated 1" 
to the right of /, 2j" above the ground, and 2" in front of the 
vertical plane. Draw the projections of the perpendicular 
from A on to each plane. 

3. The horizontal and vertical traces of a plane make angles of 30 

and 50 respectively with xy. Take a point in xy. 2" from 
/, and determine the plan and elevation of the line OM. drawn 
at right angles to meet the plane. Determine the angle which 
OM makes with xy. Ans. 62. 3 . 

4. A line AB 3" long has its ends I J" and 2^" from each of 

the planes of projection. Determine a plane which shall 
bisect AB at right angles. 

5. Draw the traces of any oblique plane. Take any two points a 

and b in the horizontal trace, and any point c' in the vertical 
trace. Find the true shape of the triangle ABC. 

6. Draw the traces of any oblique plane and those of 

parallel to it, such that the perpendicular distance 
the horizontal traces is 1". Determine the true 
between the planes. 

7. The horizontal traces of two parallel planes are \\" apart and 

make 45 with xy ; the vertical traces are ft" apart, and the 
angle between the traces k obtuse. Find the distance between 
the planes. Ans. 6". 



a plane 
between 
distance 



238 PRACTICAL SOLID GEOMETRY chap. 

204. Problem. Having given an oblique plane VTH, 
and the plan of a point A in the plane, to determine the 
projections of a straight line through A which shall lie in 
the plane, and make a given angle d with the horizontal 
trace. 

Determine ov, the edge elevation of the plane. 

Through a draw the projector to meet ov in a". Then 
a" is the auxiliary elevation of the point A. 

Obtain A , the rabatment of A about oh, as in Prob. 
1 86. Through A draw^ ^, making the given angle 8 with 
the horizontal trace oh. Then A Q b is the rabatment of the 
required line. 

When the plane goes back into its original position, the 
point B does not move, being in the axis of rotation. The 
plan of the required line is therefore ab, and the elevation 
a'b' can be drawn since the heights of A and B are known. 

205. Problem. Having given an oblique plane VTH, 
and a point A in it, to determine the projections of a 
straight line AB, which shall lie in the plane, and be in- 
clined at a given angle a to the ground. 

The line is first drawn through A parallel to the vertical 
plane ; inclined at a ; its lower end on the ground. In this 
position its projections are a'B\ and aB v 

The line is then turned about Aa until B comes into the 
horizontal trace of the plane. The line is then altogether in 
the plane; it is still inclined at a; and it passes through^. 

The required projections are ab, a'b' , the manner of 
obtaining which is obvious. Compare Prob. 191. 

206. Problem. To determine a straight line which 
shall pass through a given point P, have a given inclina- 
tion a, and be parallel to a given oblique plane VTH. 

First, by Prob. 205, draw any line AB in the plane, 
having the given inclination a. Then draw a line through 
P parallel to AB. 

The plan of the line is drawn through p, parallel to ab, 
and the elevation through p' parallel to a'b' . 



X 



THE OBLIQUE PLANE 



239 




Examples. 1. The traces vt, th of a plane make angles of 45 
and 30 with xy. Select a point a 2" to the right of / and 1" 
below xy : this is the plan of a point A in the plane. Draw 
the plan and elevation of a line AB 2^" long, which lies in 
the plane and makes 30 with the horizontal trace. 

Determine also the projections of a line AC 3" long, lying 
in the plane, the angle BAC being 50'. 

Hint. Having obtained A C , produce it to meet the 
horizontal trace in /'. To find the plan c, join ia and produce 
it to meet in c a line from C perpendicular to ///. If the point 
i obtained as above be not within the limits of the paper, 
draw 

a more convenient point i, and 
then c will lie on id produced. 

2. The traces of a plane each make 50 with xy. A point A 

in the plane is 1" from the vertical plane, and 2" above the 
ground. Determine the projections of a line AB lying in 
the plane, and inclined at 35 (a) to the horizontal plane, (b) 
to the vertical plane. What is the greatest inclination which 
AB may have either to the horizontal or vertical plane ? 

3. Determine a plane which passes through the point A, Ex. 2, 

and is parallel to the given plane. 



e other line from C fl to meet the horizontal trace in 



A B in Z) , 



from D obtain d, 



2 4 o PRACTICAL SOLID GEOMETRY chap. 



207. Problem. Having given the plan of any polygon 
or plane figure lying in a given oblique plane, to find the 
elevation and true shape of the figure. 

Draw an edge elevation of the plane as in Prob. 198, 
then proceed as in Prob. 186. 

208. Problem. Having given a straight line and an 
oblique plane, to determine (a) the trace of the line on the 
plane ; (b) the projection of the line on the plane ; and 
(c) the angle between the line and plane. 

Draw an edge elevation of the plane, and a new eleva- 
tion of the point on the same ground line. The problem 
is thus reduced to Prob. 190, and is worked in the 
manner there described. 

209. Examples on Problems 198 to 208. 

*1. Draw an edge elevation of the plane, Fig. (a), on a ground line 
taken through b. Draw also an edge plan on x-,y-., taken 
through a'. 

*2. Fig. (/>) shows the plan of a prism with equilateral ends, resting 
on the ground. Draw the elevation, and show the section by 
the given plane. Draw also a sectional projection of the prism. 

*3. Find the intersection of the line and plane in Fig. (e). Also in 

Fig. (/) 
_x "4. In Fig. (a) determine a perpendicular from A to the plane VTH. 
*5. Determine a plane bisecting the line AB, Fig. (e), at right 

angles. 
*6. Find the distance between the parallel planes of Fig. (d). Show 

the projections of any line perpendicular to the planes with one 

end in each plane. 
*7. In Fig. (a) the point B lies in the plane ; find its elevation. 

Show a line containing B, lying in the plane, and making 70" 

with the trace ///. 
*8. Determine a line through B, Fig. (a), lying in the plane and 

inclined at 30' 3 to the ground. 
*9. In Fig. (a) determine a line containing B, lying in the plane and 

making 50 with vt. Also one Inclined at 50 to the vertical 

plane. 
*10. In Fig. (<) the triangle whose plan abc is given lies in the 

given plane. Determine the elevation and true shape of ABC. 
*11. In Fig. (e) determine the trace of the given line on the given 

plane. Show the projection of the line on the plane. Find 

the angle between the line and plane. 
*12. Obtain the results of Ex. 1 1 with reference to Fig. (f). 



THE OBLIQUE PLANE 



241 




R 



242 PRACTICAL SOLID GEOMETRY chap. 

210. Problem. Having given an oblique plane VTH, 
and one projection of a point A which lies in the plane, 
to determine the other projection. 

The method of solution is to first find the projections of 
any line drawn through A and lying in the plane. Then 
to draw a projector from the given plan to intersect the 
elevation of the line in a. 

( i ) Let the plan a be given. 

In Fig. (a), through a draw be, the plan of any line 
which has its lower and upper ends, B and C, respect- 
ively in the horizontal and vertical traces of the plane ; 
the elevation of the line is evidently b'c, B and C 
being respectively in the horizontal and vertical planes of 
projection. The projector from a, intersecting b'c in a, 
gives a the required elevation. 

In Fig. (b) the line through A is horizontal, so that 
its plan ac is parallel to ///, and its elevation a'c parallel 
to xy. 

In Fig. (c) the line is taken parallel to the vertical 
plane ; therefore its plan ab is parallel to xy, and its eleva- 
tion a'b' parallel to tv. 

(2) Let the elevation a be given. 

The elevation of the line through A is first drawn ; the 
plan of the line is then determined ; and finally the 
required plan a is found by projecting from a '. 

211. Problem. Having given the plan and elevation 
of a point A, and one trace of a plane which contains A, 
to find the other trace. 

Let a, a, and th be given, Fig. (a). 

Draw be through a ; project b' ; through c draw the pro- 
jector to intersect b'a' produced in c . Then tc is the re- 
quired vertical trace. 

If the given trace be nearly parallel to xy, so that the 
point / is not readily accessible, two lines such as BC 
must be drawn through A, thus determining two points in 
the other trace. See Fig. 215. 



X 



THE OBLIQUE PLANE 



243 




y (a) {b) h 

210 and 211 




Examples. 1. Represent a plane by its traces ; select a point a any- 
where except in xy. Draw the plan of any line whatever which 
lies in the plane and passes through A. Find the elevation of 
this line, and by means of it find the elevation of A. Find the 
inclination of the line through A which you have represented. 

2. Draw a line making 40 with xy, and also the projections 

of any point whatever. Determine the vertical trace of the 
plane which contains the point, and has the first line for its 
horizontal trace. 

3. Draw a line inclined at 30 to xy, and draw the projections of 

any point not in either of the planes of projection. Determine 
the horizontal trace of the plane which contains the point and 
has the first line for its vertical trace. 

4. Draw the projections of any point. Take a point I ^". below xy, 

and through it draw a line inclined at 5 to xy. Let this be 
the horizontal trace of a plane containing the first point. 
Determine the vertical trace. 

Hint. Draw the plans of any two lines through the given 
point. Regard these as lying in the plane ; draw their elevations 
and their vertical traces, then join the latter. See Fig. 215. 
We advise the student to remember this useful construction. 

5. A point A is 1" above the ground and 1.8" behind the vertical 

plane. A second point B is 1.5" below the ground and 0.5" 
in front of the vertical plane. Determine a plane which con- 
tains A and B and is parallel to the ground line. 
6 3 Represent a plane which contains the ground line and also the 
point A of Ex. 5. Find the angle between this plane and the 
plane of Ex. 5. 



244 PRACTICAL SOLID GEOMETRY chap. 



212. Problem. To determine a plane which shall pass 
through a given point A and be parallel to a given plane 
VTH. 

The student may easily work this problem by drawing 
an edge elevation. The following is the special method : 

Through a draw ac parallel to ///. This is the plan of 
a horizontal line in the required plane, and a'c, the eleva- 
tion of the line, is drawn as shown. 

A point C in the vertical trace of the required plane 
is thus found. Through c draw Im parallel to tv, and 
through m draw mn parallel to th. The required plane is 
LMN. 

213. Problem. To find the rabatments of a given 
point A which lies in a given oblique plane VTH. 

First, to find A , the rabatment of A into the horizontal 
plane, about th. 

Draw am perpendicular to th. Then am is the plan of a 
line in the plane, at right angles to the horizontal trace. It 
is also, when produced, the plan of the circular path traced 
by A during rabatment ; so that A must lie on this line. 

Obtain a" am, the rabatment of the right-angled triangle 
MAa about am ; observe that aa" = la' . Make mA = ma", 
the true length of MA. Then A is the required rabatment 
of A into the horizontal plane. 

It will be noticed that the arc a"A is the rabatment 
about aA of the path traced by A. As before stated, 
aA is the plan of the path. 

Next, to find A v the rabatment of A into the vertical 
plane, about tv. The construction is similar to that just 
described ; an is the elevation, and a x ri the rabatment 
into the vertical plane, of a line AN lying in the plane at 
right angles to the vertical trace ; n'A 1 is the rabatment of 
the same line about tv ; and A 1 is the required rabatment 
of the point A into the vertical plane. 

A model to illustrate this important problem may be 
made in the following manner : 



THE OBLIQUE PLANE 



245 





Take a model of the oblique plane hinged along ///, and on it draw 
the line AM perpendicular to th. Cut out in paper a right-angled 
triangle of the shape a"ma, with a margin along am for attachment 
to the horizontal plane. The two rabatments about am and th re- 
spectively may then be effected. 

This model may be used to illustrate the rabatment of a point into 
the vertical plane, by turning it so that the horizontal plane becomes 
the vertical plane, and vice versa. 

Examples. 1. Draw the traces of any plane VTH, and also the 
projections of a point A lh" vertically over ht and 2" from the 
vertical plane. Determine a plane through A parallel to VTH. 

2. The traces vt, th of a plane make 50 and 6o with xy. 

Determine the rabatment, into the horizontal plane, of a point 
A in vt and 2" from /. 

3. Taking the plane of Ex. 2, determine the rabatment, into the 

vertical plane, of a point A in ht 2" from /. 

4. Determine the projections of a point A in the plane of Ex. 2, 

2" from the vertical plane and 1^" from the horizontal plane. 
Determine the two rabatments of A. 

5. A plane at right angles to xy contains a point A which is 1" 

above the ground and 2" from xy. Represent the plane and 
point and determine the two rabatments of the latter. 

6. A plane contains xy and the point -4 of Ex. 5 ; set out the two 

rabatments of A. 



246 PRACTICAL SOLID GEOMETRY chap. 

214. Problem. To determine co, the true angle 
between the traces of a given oblique plane VTH. 

Take a, a, the projections of any point A in the vertical 
trace of the given plane. 

Obtain A , the rabatment of A about th, by drawing 
aA perpendicular to ///, and making tA = ta . 

Then tA is the rabatment of the vertical trace into the 
ground, and the true angle between the traces is that 
marked w. Two cases, (a) and (b), are shown. 

215. Problem. Having given two intersecting 
straight lines, to determine the plane containing them ; 
also, to find the true angle between the lines. 

In order that the lines shall intersect, the apparent 
intersections in plan and elevation must lie on the same 
projector. 

The horizontal and vertical traces of the plane must 
pass through the corresponding traces of the line. 

Let the given lines be as shown in Fig. 215. 

As in Prob. 177, determine H, A', the horizontal 
traces, and V, S, the vertical traces of the given lines. 

Then, lines drawn through //, k and v', s' are respectively 
the horizontal and vertical traces of the required plane. 

To find the true angle between the lines, obtain A , the 
rabatment of A, as in Prob. 213. Join A /i, A k. 
Then hAJz is the rabatment of the triangle HAK, and the 
angle hA k is the true angle between the lines. 

216. Problem. Having given two parallel straight 
lines, to determine the plane containing them ; also, to 
find the true distance between the lines. 

Let the given parallel lines be as shown in Fig. 216. 
Their plans and elevations are respectively parallel. 

Determine the traces of the lines, through which draw 
the traces of the required plane, as in the last problem. 

The true distance between the lines is that between 
their rabatments, which are shown dotted in the figure, and 
are obtained as in the last problem. 



THE OBLIQUE PLANE 



247 




216 \ *T 

V * 



Examples. 1. Determine the true angle between the traces of 
each of the planes of the example on page 233. In each case 
draw the projections of the bisector of the angle between the 
traces. Also, draw the projections of a line equally inclined 
to the traces and such that the part of the line intercepted by 
the planes of projection is 3" in length. 

2. Draw the projections of any two lines AB, AC. Find (a) 

the traces of the plane containing the lines, (b) the inclina- 
tion of this plane, [c) the angle BAC, (d) the projections 
of the bisector of the angle BAC, (e) the projections of the 
perpendicular from A on to BC. 

3. Draw the projections of any two parallel lines. Determine the 

plane containing them, and the distance apart of the lines. 



248 PRACTICAL SOLID GEOMETRY chap. 



217. Problem. To determine a plane which shall 
contain three given points, A, B, C ; also, to find the true 
shape of the triangle ABC. 

Draw lines through any two pairs of the given points, 
and then find the plane containing these lines, as in 
Prob. 215. 

To find the true shape of the triangle ABC, obtain the 
rabatments of A, B, and C by the method of Prob. 213, or 
work by an auxiliary elevation, as in Prob. 207. 

218. Problem. To determine a plane which shall 
contain a given straight line and a given point ; also, to 
find the true distance of the point from the line. 

Take any point in the given line and join it to the given 
point. Then the plane through the two intersecting lines, 
found as in Prob. 215, is the one required. 

To find the true distance between the point and the 
line, rabat the plane containing them, and draw the 
perpendicular from the rabatment of the point to the 
rabatment of the line. 

219. Problem. To determine a line which shall lie 
in the plane of two given intersecting straight lines, and 
shall bisect the angle between them. 

Draw the rabatment of the lines as in Prob. 215. 

Then through A (Fig. 215) draw a line bisecting the 
angle kA k, and intersecting hk in p (not shown). 

Thus Ave obtain A p, the rabatment of the required 
bisector, and ap its plan ; and since P is on the ground 
the elevation of AP is known. 

220. Problem.- To determine the angle between two 
non-intersecting straight lines. 

Through any point in one of the lines draw a line parallel 
to the other. 

The angle between these two intersecting lines, found 
by Prob. 215, is the angle required (Def. 10, Appendix). 



THE OBLIQUE PLANE 249 



221. Problem. To find the angle between two given 
planes. 

Take any point A, and through A draw two lines 
respectively perpendicular to the two planes, in accordance 
with the theorem stated in Prob. 201. 

Find the angle between these two lines by Prob. 215. 
Then this angle, subtracted from 180, will give the angle 
between the planes. 

This solution is based on a proposition of pure solid 
geometry, which states that the angle between two planes is 
equal to the supplement of the angle between any two lines 
respectively perpendicular to the planes. Refer to Prob. 
227 for another method of solution. 

222. Problem. To find the angle between a given 
straight line and a given oblique plane. 

One method of determining this angle has already been 
given in Prob. 208. The following is an alternative 
solution : 

Take any point A in the given line, and through A draw 
a line perpendicular to the given plane. Find the angle 
between these two lines by Prob. 215. Then this angle, 
subtracted from 90, gives the angle required. 

This solution is based on a proposition of pure solid 
geometry, which states that the angle between a line and a 
plane is equal to the complement of the angle between the 
line and a line perpendicular to the plane. 

223. Problem. To detennine a plane which shall 
contain a given straight line, and be perpendicular to a 
given plane. 

Take any points in the given line, and through A draw 
a line perpendicular to the given plane (Art. 201). Then 
the plane through the two intersecting lines, determined as 
in Prob. 215, is the one required. 

This construction is based on Theorem 6 of the Ap- 
pendix, and is often required. 



250 PRACTICAL SOLID GEOMETRY chap. 



224. Problem. To determine the plan and elevation 
of the line of intersection of two given planes. 

Let VTH and LMN be the given planes ; four cases 
are shown in the figure. 

In each case let the horizontal traces intersect at r, and the 
vertical traces at/, so that RS is that part of the intersection 
intercepted by the planes of projection. 

The elevation of R is r, and the plan of *S" is s, r and s 
being in xy. Hence rs is the plan, and r s the elevation 
of the required intersection. 

In (b) the vertical traces intersect below xy, and RS is 
between the planes forming the fourth dihedral angle. In 
(d) the horizontal traces are parallel to each other, and RSis a 
horizontal line parallel to either trace. 

225. Problem. To determine the plan and elevation 
of the point of intersection of three given planes. 

The point required is that in which the line of inter- 
section of any two of the planes intersects the third plane. 
Or it may be determined as follows : 

Find RS and PQ, the lines of intersection of any two of 
the three pairs of planes. These lines, produced if necessary, 
will intersect in the required point. 

The accuracy of the work may be tested by observing 
whether the apparent intersections, in plan and elevation, lie 
on the same projector. 

Examples. 1. Take two points t, m in xy 3" apart. Above xy 
construct the triangle tlm, where // is 3j", and Im is 1.8". 
Below xy construct the triangle t?im, making tn equal 1 . 3", and 
nm 2\ ". Determine the projections of LN, the line of inter- 
section of the planes L TN and LMN. 

Find the inclinations of LN to the planes of projection. 

2. Two vertical planes meet xy at points 2" apart, and the planes 

make angles of 55 and 40 respectively with xy. Draw the 
plan and elevation of their intersection. 

3. An inclined plane makes 45 with the ground. An oblique 

plane is parallel to xy and is inclined at 50" to the ground. 
Determine the projections of the intersection of these two 
planes. 



THE OBLIQUE PLANE 



251 




4. Take any point in the intersection of Ex. 3 and draw the pro- 

jections of a line passing through this point, lying in the oblique 
plane, and making an angle of 55 with the intersection. 
Hint. Rabat the oblique plane about its horizontal trace. 

5. Draw the projections of a line AB inclined at 35 and 50 

respectively to the horizontal and vertical planes. Draw the 
traces of any pair of planes which intersect in AB. 

6. Draw the traces of any three planes and determine the projec- 

tions of their point of intersection. 



252 PRACTICAL SOLID GEOMETRY chap. 

226. Examples on Problems 210 to 225. 

*1. In Fig. (a) the plan of a point A lying in the plane VTH is 

given ; find the elevation of the point. 
*2. In Fig. (/>) the projections of a point and the horizontal trace of 

a plane are given. Find the vertical trace. 
*3. Find the traces of the three planes which pass through A, Fig. 

(e), and are parallel to the three given planes. 
*4. In Fig. (d) the point whose elevation is given lies in the given 

plane ; draw its plan and obtain the two rabatments of the 

point. Work the corresponding problems having reference to 

Figs, (a) and (/>). 
*5. Find the angles between the traces of each of the three planes 

of Fig. (e). 
*6. Find the traces of the plane determined by the two intersecting 

lines, Fig. {c). Find the angle BAC. 
*7. Find the positions of B and C on the given lines, Fig. {c), so 

that ABC shall be an isosceles triangle having AB = AC.= 2. 5". 
*8. Determine the projections of the bisector of the angle BAC in 

Fig- (<") 
*9. In Fig. (e) find each of the angles which AB makes with xy, 

with the trace LM, and with the trace MN. 
*10. Find the angles between the planes LMN and PQN in Fig. 

(e). Find also the angles between LMN and I 'TILT, and 

between PQN and VTHT. 
*11. Find the angle between the line AB and the plane LMN in 

Fig. (e). Also between AB and PQN. And between AB 

and VTHT. 
*12. Determine the three planes which contain the line AB, Fig. (e), 

and are respectively perpendicular to the planes LMN, PQN, 

VTHT. 
*13. In Fig. (e) determine the line of intersection of the planes LMN 

and PQN. Also of the planes LMN, VTHT. And of the 

planes PQN, VTHT. 
*14. Find the point common to the three planes of Fig. (e). 

15. The traces /// and vt of a plane make 55 and 45 with xy. A 

point A is in the plane and ij" above the ground. Is this 
information sufficient to define the position of A ? If not, what 
is the locus of A? Draw the projections of the locus of A. 

16. In Ex. 15 determine the positions of A, if in addition to the 

data there given, you are told (1) that the point is ii" from the 
vertical plane, or (2) 2" from xy, or (3) that the point is 3" 
from the point t where the plane cuts the ground line. 

17. Determine the intersections of the three planes of Fig. (c) with 

plane which contains xy and the point A. 



THE OBLIQUE PLANE 



^53 




254 PRACTICAL SOLID GEOMETRY chap. 

227. Problem. A given irregular triangular pyramid 
rests with one face on the ground ; to determine its six 
dihedral angles, and the true shapes of its faces. 

Let abed be the given plan of the pyramid, the face 
ABC being on the ground, and the height of D being as 
indexed. 

First to find the dihedral angle between the faces which 
intersect in AD. 

Take xy parallel to ad, and on xy draw ad', the 
elevation of AD. Draw any line vt perpendicular to a'd\ 
and /// perpendicular to ad; these lines are the traces 
of a plane perpendicular to AD. This plane intersects 
AD in P, and the two faces ADB, ADC in PR, PS, 
which two lines are perpendicular to AD {Theorem 16, 
Appendix), and therefore contain an angle which measures 
the angle between the faces in question {Definition 9, 
Appendix). 

The true angle between these lines is the angle rP s, 
obtained by a rabatment of the plane VTH about th. 

The angles between the other two pairs of sloping faces 
may be found in a similar manner. 

Next to find the angle between the faces which 
intersect in PC, and the true shape of the face DBC. 

Rabat DBC into the ground about be. The con- 
struction for this is shown in the figure ; it is the same 
as that explained in Prob. 186. The angle between the 
faces which intersect in BC is measured by the angle 
d 6 md", and the true shape of the face DBC is the triangle 
D be. 

The remaining dihedral angles, and the true shapes of 
the remaining faces, may be similarly found. 

Example. Draw a triangle ABC, making AB 3^", AC 2|", and 
CB 3|"; take a point D in the triangle if" from A and if" 
from C. Join AD, BD, CD. This is the plan of an irre- 
gular pyramid with the face ABC on the ground, and the 
vertex D if" above the ground ; determine the six dihedral 
angles and the true shapes of its faces. 



x THE OBLIQUE PLANE 255 



-?Po 



228. Some tangential properties of cones. 

Place a cone with its base on a horizontal plane, and let 
a cardboard model of a plane, having one edge on the 
horizontal plane, rest against, or be tangential to, the surface 
of the cone. It will be observed : 

(1) That the tangent plane (if sufficiently extended) 
contains the vertex of the cone ; (2) that the inclination of 
the plane is equal to the base angle of the cone; and (3) 
that the trace of the plane touches the circular base, or trace, 
of the cone. 

If the cone have its base in the vertical plane, similar 
remarks apply, having reference to the vertical plane. 

A plane can in general be found tangential to two cones. 
(1) When the cones have their axes parallel, and their 
vertical angles equal ; (2) when the cones have a common 
vertex ; (3) when the cones circumscribe the same sphere. 

If the student finds difficulty in the application of these 
principles to any of the remaining problems of this chapter, 
let him return to the problems after reading Chapters XIII. 
and XIV., where the principles are illustrated in more 
detail. See also the theorems in the Appendix. 



256 PRACTICAL SOLID GEOMETRY chap. 

229. Problem. To determine the inclinations e and 
9 of a given oblique plane to the horizontal and vertical 
planes of projection. 

Method i. Suppose a cone with its base angle equal to 
6, to be cut into two halves by a plane containing the axis ; 
then one of these half-cones may be placed with its 
triangular face in the vertical plane, and the semicircular 
base on the ground, so that the given plane shall be 
tangential to it. The following solution consists in 
drawing the plan and elevation of the half- cone in this 
position, and so finding the base angle 0. 

Take for the vertex of the half-cone any point S in the 
vertical trace, s, s being its projections. With centre .? describe 
the semicircle man touching th at o. This is the plan of 
the half-cone. Join / to m and n. Then tns'n is the 
elevation of the half-cone, and the base angle marked 6 is 
the required inclination of the plane to the ground. 

To find <f>, take for the vertex of a half-cone any point R in 
the horizontal trace, r, r being its projections. With centre 
r, describe the semicircle pk'q, touching tv at k' . This 
semicircle is the elevation of a half-cone with its base in the 
vertical plane, and to which the plane is tangential. The 
plan of the half-cone is rpq, and the base angle, marked c/>, 
is the required inclination of the plane to the vertical plane. 

Method 2, The inclination 6 may also be found by 
drawing an edge elevation of the plane as in Prob. 198. 
The inclination </> may similarly be obtained by drawing an 
edge plan of the plane on x x y v taken perpendicular to the 
vertical trace. 

The model illustrating the first of these constructions will also serve 
to illustrate the second, by turning it into a position so as to reverse 
the planes of projection. And the construction given for the first will 
also serve to illustrate the second, if it be read with the page upside 
down, so as to reverse the traces of the plane. 

Method 3. The construction for the rabatments of a 
point A, in Prob. 213, also gives the inclinations of the 
plane, the angle 6 being equal to the angle ama" of Fig. 
213, and the angle 4> equal to ana l of the same figure. 



THE OBLIQUE PLANE 



257 



&V 




230. Problem. Having given one trace of a plane, and 
also the inclination of the plane to one of the planes of 
projection, to find the other trace. 

Let the horizontal trace th be given, and also 8, the 
inclination of the plane to the ground. 

Take any point j - in xy, and with centre .f describe the semi- 
circle tfion, touching the given horizontal trace at 0. Through 
m and u draw lines making with xy, and intersecting 
at s\ Then the line ts' is the required vertical trace. 

Examples. 1. Determine the inclinations 6 and $ of a plane whose 
horizontal and vertical traces make 30 and 50 with xy. 

2. A line intersecting xy at 50 represents both traces of an 

oblique plane. Find 9 and <p, the inclinations of the plane 
to the horizontal and vertical planes. 

3. The real angle between the traces of a plane inclined at 50 is 

8o c ; represent the plane by its traces. 

4. A line making 40 with xy is the horizontal trace of a plane 

inclined at 50 to the horizontal plane ; determine its vertical 
trace. 

5. A line making 45 with xy is the vertical trace of a plane inclined 

at 60 to the ground. Determine the horizontal trace. 

6. A line making 6o with xy is the horizontal trace of a plane 

inclined at 65 to the vertical plane. Determine its vertical 
trace. 

7. A line making 40 with xy is the vertical trace of a plane 

inclined at 6o to the vertical plane. Determine its horizontal 
trace. 

S 



258 PRACTICAL SOLID GEOMETRY chap. 

231. Problem. To determine the traces of a plane 
which shall be inclined at given angles e and 9 to the 
horizontal and vertical planes of projection. 

First observe, as stated in Art. 228, that if there be two 
cones such that the vertex of the second cone does not fall 
within the first cone, we can conceive a plane which touches 
the first cone and passes through the vertex of the second. 
Now this plane will also touch the second cone if the two 
cones circumscribe the same sphere. See the theorems of the 
Appendix, relating to the sphere, cone, and cylinder. 

With any centre c in xy, and any radius, describe a 
circle. This is both plan and elevation of a sphere with 
its centre in xy. 

Draw tangents v'r , v's to this circle as shown, each in- 
clined at to xy ; and with centre c describe the semi- 
circle r'as'. These are the projections of an upright half- 
cone, base angle 0, circumscribing the sphere. 

Also draw tangents v x e, v x f\o the circle, each making <f> 
with xy, and draw the semicircle ea^f as shown. These 
are the projections of a second half-cone circumscribing the 
sphere, with its base in the vertical plane, and base angle 
equal to </>. 

Finally, from v' draw the tangent v'a^ to the semicircle 
ea-[f\ and from v x draw the tangent v x a to the semicircle 
r'as' . These tangents will be found to intersect in a point 
/ in xy, and are the traces of the plane required. 

The projections of the two generators A V, A x V v and the 
point P in which the plane touches the two cones and 
sphere are shown in the figure. 

Limits to the data. A tangent could not be drawn from v to the 
semicircle on ef, if v' were inside this semicircle ; therefore, cv' must 
not be less than ce. Now, observing that cov' and co x e are right-angled 
triangles, having the sides co, co. equal, it appears that if cv is greater 
than ce, the angle ov'c opposite oc must be less than <p opposite o-^c. 
But 6 + ov'c = go, therefore d + (p cannot be less than 90. Since 
neither nor (p can be greater than 90 , d + (p cannot be greater than 
180 . 

Thus 6 + <P must lie between 90 and 180 , both inclusive. 



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Examples. 1. Determine the traces of a plane inclined at 6o 
to the ground, and 45 to the vertical plane. 

2. A point A is 2j" above the ground, and i|" in front of the 

vertical plane ; determine the traces of a plane passing through 
A and inclined at 40 and 55 to the horizontal and vertical 
planes of projection. 

3. Draw the traces of a plane which is inclined at 45 and 50 to 

the horizontal and vertical planes of projections, the angle 
between the traces being obtuse. 

Hint. Take the vertical cone in the above figure, with its 
apex downwards. 

4. Determine two planes, one of which is inclined at 40 and 50 

to the horizontal and vertical planes of projection, and the other 
at 90 and 90 . Find the angle between these planes. Ans. 
90. 

5. Work Prob. 231 when = 6o, and ^=45, taking the radius 

of the sphere to be 1". Determine the contact of the plane 
with each cone and with the sphere. 



260 PRACTICAL SOLID GEOMETRY chap. 

232. Problem. -To determine a plane which shall be 
inclined at a given angle e, and shall contain a given 
straight line AB. 

Find Hand V, the horizontal and vertical traces of the 
line. These are points in the traces of the required plane. 

Next represent a cone with its base on the ground, its 
vertex at any point in the line, say A, and having a base 
angle equal to 6. The required plane will be tangential to 
this cone, and the horizontal trace must touch the plan 
of the circular base (Art. 228). 

The isosceles triangle with its vertex at a, and its 
base angle equal to 0, is the elevation of this cone ; the 
plan is the circle with centre a. 

The horizontal trace of the required plane is the line 
ht drawn through h to touch this circle, and the vertical 
trace is the line drawn through / and v', where v is the 
elevation of the vertical trace of BA. 

233. Problem. To determine the traces of a plane 
which shall be perpendicular to a given oblique plane 
VTH, pass through a given point A, and have a given 
inclination e. 

The required plane will be perpendicular to the plane 
VTH, if it contain any line perpendicular to the plane 
VTH 

Therefore draw ar perpendicular to ht, and dr perpen- 
dicular to vt ; these are the projections of a line AR 
perpendicular to the plane vth. The problem therefore 
reduces to the last one. 

Therefore draw the projections of an upright cone 
with vertex A and base angle 9. Obtain n, the horizontal 
trace of AR ; and from n draw a tangent (there are two) 
to the plan of the base of the cone, thus obtaining mn, 
the horizontal trace of a plane satisfying the required 
conditions. The vertical trace ml is readily obtained 
since A is a point on the required plane. 

For examples see page 263. 



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262 PRACTICAL SOLID GEOMETRY chap. 

234. Problem. To determine the traces of a plane 
passing through a given point A, having a given inclina- 
tion o, and making an angle a with a given plane VTH. 

If there be two cones having ias a common vertex, 
their base angles being and a respectively, then if these 
cones have their bases on the ground and the given plane 
respectively, a plane which touches both cones will satisfy 
the given conditions. 

Convert the given oblique plane VTH into the inclined 
plane VOH as in Prob. 198, and obtain a", the new 
elevation of A on x'y. 

Now draw d'r" s" and a"u"w", the elevations (on x'y) of 
the cones described above. 

The horizontal trace of one cone is the circle with 
centre <?, and that of the other is an ellipse whose elevation 
is ef. 

We shall avoid having to draw this ellipse in plan, by 
the following artifice : 

Any convenient sphere is conceived as inscribed in the 
cone whose elevation is a"e"f" ; c" is the elevation of the 
centre of such a sphere. This sphere is then conceived as 
circumscribed by a cone with its base on the ground, and 
base angle 6 ; the plan and elevation of this cone are 
shown. Observe that the plan c of the centre of the base 
of this last cone will be in the line through a at right angles 
to ///. 

Draw a common tangent nm to the plans of the two 
upright cones. This is the horizontal trace of one plane 
satisfying the given conditions, and the vertical trace 1m 
is readily obtained as shown, since A is in the plane. 

For, a plane which touches the two vertical cones has 
the given inclination ; it also passes through the given point 
A, and touches the sphere centre C, and therefore is 
tangential to the cone whose elevation is a"u"w". 

Note 1. The trace mn might have been obtained by first draw- 
ing the elliptical trace referred to above, and then drawing 
a common tangent to this ellipse, and the circle centre a. 



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263 




3. 
4. 



Examples. 1. A point A is 2" in front of the vertical plane, and 
I A" above the ground. A point B is 1^" to the left of A, \" 
below it, and J" farther from the vertical plane. Determine the 
traces of two planes each containing AB and inclined at 50 to 
the ground. 

2. The traces vt, th of a plane make angles of 50 and 45 
with xy. A point A is iA/' to the right of /, 2" above the 
ground, and iA" in front of the vertical plane. Determine the 
traces of a plane which passes through A, is inclined at 65 
to the ground, and is perpendicular to the given plane. 
Determine a plane inclined at 65 , passing through the point 
A, and making an angle of 6o with the plane in Ex. 2. 
A plane parallel to xy is inclined at 6o. Determine a second 
plane which shall be inclined at 45 both to the first plane and 
to the vertical plane. 



264 PRACTICAL SOLID GEOMETRY chap. 

235. Problem. To determine the traces of a plane 
which shall pass through a given point A, make a given 
angle a with a given line AB, and have a given inclina- 
tion e. 

First observe that the required plane will touch each of 
two cones, namely : 

(1) An upright cone with vertex A, and base angle 9. 

(2) A cone with vertex A, axis AB, and semi-vertical 
angle a. 

The projections of the first cone are easily drawn, and 
those of the latter may be obtained in the following 
manner : 

Suppose AB to turn about the vertical line Aa until it 
is parallel to the vertical plane. The plan of AB will now 
be aB v and the elevation a'B{. 

Through a draw two lines each making an angle a with 
a'B^, and describe a circle to touch these lines, the centre 
being any point c^ on a B^. 

Now describe a circle with the same radius, but with c 
as centre, qV being parallel to xy. Draw tangents to this 
circle from a, then these form the elevation of a cone of 
indefinite length, having AB for its axis and a semi-vertical 
angle a, C being the centre of an inscribed sphere. The 
outline of the plan of this cone is not required. 

Next draw the projections of an upright cone with base 
angle 9 circumscribing the sphere, centre C, its base being 
on the ground. 

Draw nm to touch the plans of the bases of the two ver- 
tical cones, then nm is the horizontal trace of the required 
plane (there are two such), and the vertical trace is readily 
drawn because A is a point in the plane. 

It will be obvious that since the plane LMN touches 
the two upright cones, it must pass through A and touch 
the sphere centre C ; hence it must touch the inclined 
cone. 

Note. In connection with this problem the student should study 
the theorems on the cone and cylinder given in the Appendix. 



THE OBLIQUE PLANE 



26: 




Examples. 1. A line AB is parallel to the vertical plane and 
distant i|" therefrom ; its inclination to the ground is 50 . De- 
termine the traces of a plane which makes 30 with AB and has 
an inclination of 6o. 

Note. Observe that there are limits to the data in this 
problem. Thus the inclinations of the line and plane being 
50 and 6o, show that the angle between the line and plane 
may not be any angle, but must be between o and 70. 

2. Find the angle between a face of a cube and a diagonal of the 
solid. Then determine a plane which contains the face and is 
inclined at 65 , the diagonal of the cube being parallel to the 
vertical plane and inclined at 45 . 



266 PRACTICAL SOLID GEOMETRY chap. 

236. Problem. The projections of a line AB being 
given, and the horizontal trace, ht, of a plane which passes 
through A, it is required to determine the projections of a 
line AP lying in the plane HT and making a given angle 
6 with the given line. 

Note. The given angle must not be less than that between the 
given line and plane. 

First determine and set out the true length of AB. 
Draw AP so that the angle BAP is [3, and choose any point 
C on AP. 

We can now find the projections of C because it lies on 
the surface of 

i. A sphere, centre A, radius A C. 

2. A sphere, centre B, radius BC. 

3. The given plane. 

Draw x'y at right angles to ht ; project a", then oa" is 
the edge view of the given plane. Project b" . 

With a as centre and AB as radius describe a circle. 
With b" as centre and BC as radius describe another circle. 
These are the elevations, on x'y', of the two spheres men- 
tioned above, and they are intersected by the given plane 
in circles the elevations of whose diameters are s"w" and 
r u . 

Rabat the given plane and these two circles into the 
ground as shown ; the rabatments of the circles intersect in 
C (there are two such points). 

From C obtain C ' and c"; from C" and c" obtain the 
plan c, then ac is the plane of one line fulfilling the required 
conditions. 

The elevation of AC on xy can be drawn since the 
height of C is known. 

We have given the above method because the device of 
employing spheres to fix the position of a point is one 
which may be used in solving many problems. 



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267 




Examples. 1. A line AB is parallel to the vertical plane and 
inclined at 6o to the ground. An inclined plane passes through 
A and makes 70 with the ground. Determine the projections 
of a line AC 2" long which lies in the given plane and makes 
65 with AB. 

Would the solution be possible if 45 were substituted for 
65 ? Ans. No. 

2. A line AB makes 30" and 50 with horizontal and vertical 
planes; draw its projection if.-/ is ih" from each plane. A 
plane contains A and makes 45 and"55 with the horizontal 
and vertical planes ; draw its traces. Determine the projections 
of a line AC 2" long which lies in the plane and makes 60 
with AB. 
3. The traces vt, th of a plane make 50 and 35 with Ay. Deter- 
mine a line which passes through 7\ lies in the plane VTH, 
and makes an angle of 65 with the ground line xy. 



268 PRACTICAL SOLID GEOMETRY chap. 

237. Problem. An oblique plane VTH is given, and the 
plan of a straight line CD which lies in this plane. It is 
required to determine a plane which contains CD and 
makes an angle a with the given plane. 

Convert the oblique plane VTHmto the inclined plane 
V'OH by Prob. 198, and project d" from d. 

Draw d"e" perpendicular to ov' . Draw the elevation, 
d"a"r", of a cone which has DE for its axis, and semi-verti- 
cal angle equal to 90 - a. 

By Prob. 274 determine aa v bb v f,f v the plans of the 
axes and foci of the ellipse (but not the curve) in which this 
cone intersects the horizontal plane. 

Now the required plane must touch this cone, hence its 
horizontal trace must touch the ellipse; it must also pass 
through c, the horizontal trace of CD. 

Therefore by Prob. 94 draw a tangent from c to the 
ellipse ; this is ;/;;/. And ml drawn through d' will be the 
required vertical trace. 

Thus Imn represents the required plane. 

Example. The traces vt, th of a plane make angles of 45 and 
35" with xy. C and D are points in the traces, each distant 
2.5" from t. Determine a plane which contains CD and makes 
an angle of 50 with the plane vth. 

238. Examples on Problems 225 to 237. 

1. A point is ij" from xy. State and project its locus. Ans. A 

cylinder of indefinite length. The projections are straight lines 
distant Ij" from, and parallel to xy, 

2. A point is 1^" distant from a point c in xy. State and project 

its locus. Ans. A sphere 3" diameter with centre at c. The 
projections are coincident circles. 

3. A point is 1" above the ground. State and project its locus. 

Ans. A horizontal plane 1" high. The vertical trace is parallel 
to xy. 

4. A point is 1 J" from xy, and 1^" distant from the point c in xy. 

State and project its locus. Ans. The circle in which the 
cylinder and sphere of Exs. I and 2 intersect. Projections, 
coincident straight lines perpendicular to xy. 

5. A point is Ij" from xy and 1" above the ground. State and 

project its locus. Ans. The two horizontal lines in which the 
cylinder of Ex. 1 is cut by the plane of Ex. 3. 



X 



TFIE OBLIQUE PLANE 



269 




6. A point is distant l|" from the point c in xy, and 1" above the 

ground. State and project its locus. Ans. The horizontal 
circle in which the sphere of Ex. 2 is cut by the plane of Ex. 3. 

7. A point is I J" from xy, ij" from a point c in xy, and 1" above 

the ground. Determine its position. Ans. Either of the two 
points common to all the loci of Exs. 1 to 6. 

8. The traces of th and tv of a plane make 55 and 45 with xy. A 

point A is 2" distant from t, i\" from the plane VTH, and 
1 y from the vertical plane. Find all the positions of A which 
satisfy these conditions. 

9. Three planes are mutually perpendicular. One is inclined at 

40 , a second at 6o ; find the inclination of the third. 

10. Three lines are mutually perpendicular. One is inclined at 40 , 

a second at 30 ; find the inclination of the third. 

11. Draw the complete plan of an equilateral triangle ABC of 3" 

edge, having given the plan ab is 2\" long and the plan ac 
makes 30 with ab. 



270 PRACTICAL SOLID GEOMETRY chap. 

239. Miscellaneous Examples. 

*1. Determine the true angle between the lines AB and BC. 

*2. Determine the projections of a circle of l|" diameter lying in the 
plane containing the two given lines AB, BC, produced if 
necessary, and touching both. ('891) 

*3. The traces of a plane are given and the projections of a point P. 
From P draw two lines each 2j" long, meeting the plane in 
points on the same horizontal line and 2^" apart. (1890) 

*4. From the given points A, B, draw two lines meeting on the 
given plane and making equal angles with it. (1891) 

*5. A /' are the projections of a given point P ; ab, a'b' those of a 
given line AB. Draw the projections of an equilateral triangle, 
with one vertex at P and the side opposite that vertex on the 
line AB. (1897) 

*6. a is the plan of a point A lying in the given plane VTH. 

Through A draw a line in the given plane VTH parallel to the 

given plane BOC. (1888) 

" *7. Draw a plane perpendicular to the plane of the triangle ABC 

and bisecting the sides BC and AB. ('894) 

8. Draw the traces of any plane equally inclined to the planes of 

projection, and determine the true angle between its traces. 

9. A plane is equally inclined to both co-ordinate planes, and the 

real angle between the traces is 50. Draw the traces. (1892) 

10. The vertical trace of a plane makes 40 with the ground line, 

and the plane is inclined at 50 to the horizontal plane. Draw 
its horizontal trace. Determine the point (P) in the plane 1" 
in front of the vertical plane and I h" above the horizontal plane, 
and through P draw a line in the plane making equal angles 
with its traces. (1889) 

11. Draw a plane inclined at 45 to the horizontal plane, and at 6o 

to the vertical plane. Draw a line in the plane inclined at 30 , 
and 2" long between its traces. Lastly, draw the plan of a 
regular hexagon lying in the plane of which the above line is a 
diagonal. ( 1897) 

12. Draw the traces of any plane inclined at 40 . Determine the 

projections of a line in this plane inclined at 27 . Draw the 
traces of a plane containing this line, and inclined 63 . Find 
the angle contained by these two planes. Determine also the 
perpendicular distance between xy and the intersection of the 
two planes. 

13. The traces vt, th of an oblique plane make 30 and 50 with 

xy. Take a point a 1" from ht and 2" from xy ; this is the 
plan of a point A 2^" above the ground. Draw af 2J" long 
making 15 with xy. Determine the height of F (1) if AF 
is 3-n" long, and F is lower than A ; (2) if AF meets xy. 



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271 



Cofiy /tie figures double size. 7? > 




272 PRACTICAL SOLID GEOMETRY chaf. 



*14. ad is the plan of a line in a plane whose traces are ab, be. One 
side of an equilateral triangle of 1.5" side lies along ad, one 
extremity at a. The plane of the triangle is inclined at 40 to 
the horizontal plane. Draw the projections of the triangle. 

(1895) 

" ;f 15. A line AB and a plane are given. Through AB draw a plane 
perpendicular to the given plane and determine the line bisect- 
ing the angle between AB and the line in which the planes in- 
tersect. (1892) 

*16. ABC is the base of a tetrahedron, and is in the horizontal 
plane. D is at 2.5" above it. (a) Find the values of the 
dihedral and plane angles round the vertex D. (b) Pis a point 
in the base ; find the points where perpendiculars from P meet 
the three other faces of the tetrahedron. ^895) 

*17. Determine the traces of any two planes containing the given 
line AB and including an angle of 6o. (1884) 

*18. / is the plan of a point P distant h" (measured perpendicularly 
to the surface) from the plane VTH and above it. Through 
P draw 

(1) a line parallel to the plane VTH, and inclined at 45 to 

the horizontal plane ; 

(2) a line also parallel to the plane VTH, but making an angle 

of I 5 with the line inclined at 45. (1896) 

*19. AB is a given line, C a given point. Find the scale of slope of 
the plane of A, B, and C ; and draw the plan of a square in that 
plane, one diagonal to be a perpendicular let fall from C on the 
line .-^9. Unit = o.i". (1895) 

*20. Determine a line through the point C to meet the line AB at a 
point D such that the angle CDA shall be equal to 50 . 
Unit = 0.1". (1894) 

21. Draw an arc of a circle 3" in diameter with centre v. Along 
the circumference set off adjacent chords ab, be, 2" and 1^" 
long. Join av, bv, cv. Assuming the circle to be in the hori- 
zontal plane and the height of V to be 3" above this plane, de- 
termine the angle made by the plane ABV, with the plane 
CBV. (1893) 

*22. Draw (1) a plane (M) inclined at 65 to the horizontal plane, 
and containing AB ; (2) a plane (7V) through B inclined at 
50 to the horizontal plane, and perpendicular to M. (1896) 
23. The horizontal and vertical traces of a plane make angles of 35 
and 40 with xy. Draw the plan of any line lying in this 
plane and inclined 30 . Now draw the plan of a line which is 
parallel to this, and lies in the plane, so that the part of it lying 
between the traces of the plane is 2". 



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273 



Copy the figures double size. 




o 



12 -5 



CHAPTER XI 

HORIZONTAL PROJECTION, OR FIGURED PLANS 

240. Problem. Having given the indexed plans of 
three points A, B, C, to determine the indexed plan of a 
horizontal line which passes through A and intersects BC. 

Case I. Let a s , b vi , c 6 be the given plans. 

First Method, Fig. (i). Join b 12 c 6 , and draw b v2 b' and 
c t .c perpendicular to b l2 c 6 , and equal to 12 and 6 units 
respectively 5 join b'c . 

Make b V2 h equal to 8 units, and draw /id', d'd respect- 
ively parallel and perpendicular to b r / G ; affix the suffix 
8 to the point d, then a s d $ is the required plan. 

Second Method, Fig. (2). Draw any xy which is not 
perpendicular to b v> e 6 , nor so nearly perpendicular as to 
lead to an ill-conditioned construction, and obtain a, b', c 
the elevations of A, B, C. 

Join b'c, and draw ad' parallel to xy ; draw the projector 
d'd; affix the suffix 8 to d ; then a s d 8 is the required plan. 

Note. By this method xy may be drawn with the tee-square, and 
the set-square used for the projectors. 

Case II. Let a v Ik, c b be the given plans, Fig. (3). 
The line BC is nearly horizontal, and the point on it at the 
level of A is supposed to be inaccessible. 

Determine the elevations of A, B, C as in the second 
method of Case I. Join a'b' ; draw e'e parallel to xy ; pro- 
ject / to e 5 ; join e 5 c 5 , and draw a Y d x parallel to e 5 c 5 . Then 
a x d x is the required plan. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 275 




Unit = 075" 



241. Problem. To determine the scale of slope of a 
plane containing three points A, B, C, the indexed plans of 
which are given. 

Let g 8 , ?> 10 , <r 6 be the given indexed points, Fig. (2). 

By one of the methods of Prob. 240 determine a^d^, 
the plan of a horizontal line in the plane of ABC. 

Draw any double line at right angles to d s a s . Let d & a 8 
and a line through b 12 , parallel to d s a 8 , meet this double 
line in gmf Then the requited scale of slope is g s f lT 

Example. Draw a triangle 20 Vi5' making a%p % 2^", a 2( f yo 3", and 
8 f 15 3|". Draw the plan of a horizontal line CD which inter- 
sects AB. Draw an elevation of ABC on a plane at right 
angles to CD. Draw a scale of slope for the plane of ABC. 



276 PRACTICAL SOLID GEOMETRY chap. 

242. Problem. Having given a plane by its scale of 
slope ab, to determine (1) the inclination e of the plane ; 
(2) a scale of slope of the plane which passes through the 
given point P and is parallel to the given plane ; and (3) a 
scale of slope of the plane which passes through P, is per- 
pendicular to the given plane, and is inclined at a given 
angle 9. 

(1) Draw xy parallel to ab, and obtain the elevation a'b' 
by making nd =10 units, and mb' = 30 units ; join b'd, and 
produce to meet xy. Then a'b' is an edge view of the 
plane, and the angle marked 9 is the required inclination. 

(2) Obtain p', the elevation of P, and draw p'e parallel 
to b'd ; then p'e is an edge view of the required plane, and 
e'e drawn perpendicular to xy is its horizontal trace. 

At any point e in this trace draw the double line parallel 
to ab, and on it mark off ed equal to, say, 30 units measured 
from the scale of ab ; index the two points of the scale o and 
30 as shown, and the scale of slope of the required plane is 
completed. Observe that since the planes represented by ab 
and ed are parallel, equal differences of level of the two 
planes correspond to equal lengths on their scales of 
slope. 

(3) Every plane passing through P at right angles to the 
given plane ab must contain the line through PaX. right angles 
to that plane. 

Draw p'Ji perpendicular to db', and P u b Q parallel to ab ; 
then these are the projections of the line through P per- 
pendicular to the given plane, and H is the horizontal trace ; 
hence the horizontal trace of the required plane will pass 
through Ary. 

Draw the projections of a cone with its base on the 
ground, its base angle <, and its vertex at P ; the required 
plane will touch this cone, and the horizontal trace will 
touch the plan of the base of the cone. 

The plan of the cone is the circle, centre p n ; hence the 
horizontal trace of the required plane is the tangent h t 
drawn from /i Q to this circle. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 277 




Unit = .031 



Take any point/ in // t ; draw the double line^ r perpen- 
dicular to //,/, and draw pg parallel to h Q t ; then ft is the 
horizontal trace of the required plane, and p 41 g is the plan 
of a horizontal line at the level 41 ; hence the scale of slope 
fg can be indexed as shown. 



Examples. 1. Draw a double line ab 2" long, and attach indices 
of 10 and 30 to a and b. Regard this as the scale of slope of a 
plane. A point c 95 is 3" from a, and 2^ from b. Determine 
the scale of slope of a plane (a) through C parallel to the given 
plane ; (b) through C perpendicular to the plane and inclined 
at 65 to the ground. Unit = o.i". 

2. Determine the plan of a line in the plane (a) Ex. 1, at a level 15, 
and of one at the same level in the plane (/') ; show the plan of 
the intersection of these lines, and index it. 



278 PRACTICAL SOLID GEOMETRY chap. 

243. Problem. The figured plan c 25 d 13 of a line, and 
the scale of slope ab of a plane are given ; to determine 
(1) the figured plan of the point of intersection of the line 
and plane ; (2) the angle between the line and plane ; and 
(3) the bisector of this angle. Unit OT'. 

(1) Take ab as a ground line, and obtain ab' the edge 
view of the given plane, and c'd! the elevation of the given 
line CD ; these intersect in /'. Draw the projector i'i, 
then i is the plan of the point of intersection of the given 
line and plane, and its index may be obtained by measuring 
i' in. 

(2) Determine en', e 2f n, the projections of CJV, the 
perpendicular from C to the plane. Then obtain Z JV Q by 
rabatment of the plane about its horizontal trace ; and next 
find C by rabatment of the triangle INC about I Q N . 

Then JV I Q C is the required angle between the line 
and plane. 

(3) Bisect the angle JV f C by I^H^. This is the 
rabatment of the required bisector. 

Set off n'li! equal to N^H^ and project h from ft. Then 
ih is the plan of the bisector. The indices for i and h 
may be found by measuring the heights of i' and ft above 
ab. 

Example. Draw a triangle abc, making ab 2", ac=$", and 
be = 2.5". Take d 2" from b and if" from c, and outside the 
triangle abc. Index the points a and d, 8 and 25 ; this is the 
plan of AD. Regard be as the scale of slope of a plane, B 
and C being at levels 10 and 30. Unit o. 1". 

(a) Determine the indexed plan of the point of intersection 

of the given line and plane. 

(b) Find also the angle between the line and plane. 

(c) Draw the indexed plan of the bisector of the angle {!>). 

(d) Draw the scale of slope of a plane which bisects the line 

AD at right angles. 

(e) Determine a scale of slope of the plane which contains 

AD and is parallel to BC, and of the plane which con- 
tains BC and is parallel to AD. Find the distance 
between these parallel planes. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 279 




c 



2S0 PRACTICAL SOLID GEOMETRY chap. 

241 Problem. To determine the line of intersection 
of two planes given by their scales of slope, ab and cd. 

Case I. The method here employed depends upon the 
fact that two horizontal lines at the same level, and one 
on each plane, must intersect each other at a point which 
lies in both planes, that is, on their intersection. 

Through the division o on each scale draw lines aj, cj 
respectively perpendicular to the scales of slope ; these are 
the plans of two horizontal lines at a level o, one on each 
plane, hence j is the indexed plan of a point on the line of 
intersection. Similarly, bi and dl drawn through the divi- 
sions 10 determine z 10 , the plan of another point on the line 
of intersection ; hence If is the required intersection. 

Case II. Let the scales of slope be nearly parallel ; 
the above method will be inconvenient. 

Conceive both planes to be cut by any vertical plane the 
plan of which is, say, /;;/. Draw the plans of the horizontal 
lines through a and b on one of the given planes, and 
through c and d on the other, intersecting hn in g, h, e',f 
respectively. Obtain g'ti and e'f , the elevations of GHand 
EF, on //;/ as ground line ; they intersect in i, from which 
the plan i, in /;//, is determined by projecting from i' . Now 
the point / is in the plane ab, since it is in the line GH 
contained by this plane ; and / is in the plane cd, since it 
is in IE; therefore lis in both planes, or is one point in 
their required intersection. On measuring W the index for 
i is seen to be 7.5. 

By taking any other vertical plane, say no, the point j 8 
is obtained in a similar manner ; hence / 7 . 5 / s is the required 
indexed plan. 

This method is applicable to any case where the line of 
intersection is within the limits of the paper. 

Case III. (no figure). Let the scales of slope be 
parallel to each other. In this case the horizontal lines 
on the two planes are parallel to each other, and the line 
of intersection of the planes is horizontal. We may proceed 
as in II., or as follows : 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 281 




Take xy parallel to ab or cd, and obtain a'b\ c'd', the edge 
views of the two planes ; let these intersect in /'. Then i' is 
the end view of the line of intersection, and the plan ij may 
be obtained by a projector from 1. The distance of i' from 
xy determines the indices of i and j. 

Example. Draw a quadrilateral abed, making ab = bc=z", ad= 
2tt", abc=jo, bad =6o. Regard a. M d and b. l; c w as the scales 
of slope of two planes ; determine the indexed plan of their 
line of inteisection. Unit 0. 1". 



282 PRACTICAL SOLID GEOMETRY chap. 

245. Problem. To determine the scale of slope of the 
plane which bisects at right angles the angle between two 
lines AB, AC, whose indexed plans are given. 

The required plane must contain the line which bisects 
the angle BAC. It must also contain the line through 
A at right angles to the Diane of BAC. These two lines 
define the plane, and their projections are first found. 

On b 30 a w produced determine n v the plan of a point N 
on the same level as c, by the method of Prob. 240 ; 
join c x n v Then CN is a horizontal line in the plane of 
ABC. 

Draw xy at right angles to cn x , and obtain cab', the 
edge elevation of the triangle CAB. Now conceive that 
the triangle revolves about CN until it is horizontal ; its 
plan while in this position may be obtained thus : 

Through c draw w'z' parallel to xy, and with c as 
centre describe the arcs b'B ' and a'A ' ; draw bB and 
and aA parallel to xy to meet projectors from B ' and A Q ' 
in B and A . Then A B c i is the required plan. 

Bisect the angle B A c i by the line A D . Find d', the 
elevation of D when the triangle ABC is brought back 
to its original position ; also, from d' project d 17 on b S0 c 4 , 
measuring the index 1 7 from the elevation. 

Through a draw a'h' perpendicular to cab' ; this is the 
elevation of the line at right angles to the plane of the 
triangle, and a 10 /i Q drawn at right angles to n^ is its plan. 

The required plane contains AD and AH, and its 
horizontal trace could be determined by obtaining the 
horizontal traces of AD and AH, and then joining these ; 
the scale of slope would be at right angles to the joining 
line, and could be indexed, since the height of a point A in 
the plane is known. 

In the figure the horizontal trace of AD is not found. 
We avail ourselves of the fact that a line which inter- 
sects AD and AH must lie in the required plane. 

Draw d'e parallel to xy ; this is the elevation of a hori- 
zontal line intersecting AD and AH; its plan d l7 e l7 is 



:i HORIZONTAL PROJECTION, OR FIGURED PLANS 283 




obtained by projecting from d' and e on to b m c< and h Q a 10 , 
and measuring and indexing the level 17. 

Draw the scale of slope at right angles to e^d^, meeting 
the latter in s, and draw h Q r parallel to e 17 d ir to meet the 
scale of slope in r. Then rs, with the indices o and 1 7, is 
the required scale of slope. 

Example. Determine the scale of slope of the plane which bisects 
at right angles the angle BAC in the example page 275. Also 
that for the angle ABC. 



284 TRACTICAL SOLID GEOMETRY chap. 

246. Problem. To determine the angle between two 
planes given by their scales of slope ab and cd. 

Determine L J the plan of the intersection of the planes, 
Prob. 244. 

Let a plane be taken at right angles to the intersection, 
cutting the given planes in two lines and the horizontal 
plane in a third line ; these lines form a triangle with its 
base on the ground, the vertical angle being the one 
required. 

Obtain i'j the elevation of the line of intersection, on 
/' j as ground line. Draw vt, th the traces of an inclined 
plane at right angles to IJ. Let vt cut j Q t" in r '. Draw 
/ a, j c perpendicular to ab, cd, and produce these lines to 
meet h Q t in h , f . 

Now rabat the triangle FRH about its base ; that is, 
with centre /, draw the arc r R w and join /i R , f R - We 
thus obtain f R /i Q , the required angle between the planes. 

247. Problem. To determine the scale of slope of the 
plane which bisects the angle between two planes given 
by their scales of slope ab, cd. 

Determine f R Q /i , the angle between the given planes in 
the manner explained in Prob. 246, and draw ^ bisecting 
the angle f R Q /i , and meeting f Q /z in g . 

Now the triangle f R /i is the rabatment of the triangle 
FRH about FH, and R g is the rabatment of the bisector 
RG. The required plane will bisect the plane angle FRH, 
and will therefore contain RG. 

But since G is on FH it will not move during the rota- 
tion of the triangle FRH, hence the required plane con- 
tains g , which coincides with G. It also contains JI, the 
intersection of the given planes. 

Join Jq^q- Then since j is the horizontal trace of JI, 
J0S0 wl ^ b e tne horizontal trace of the required plane. 

Draw the double-lined scale of slope at right angles to 
g j and meeting it in s; draw / 20 / parallel to ( ^; index s and /, 
o and 20 respectively, then si is the required scale of slope. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 285 




246 and 247 



Examples. 1. Draw a quadrilateral abed having given sides ab 
4.5", /v = 4.o", ^=2.5", da = 2.0" ; diagonal ^=4.0". -Let 
the indices of the points a, b, c be 10, 30, 5. Unit = o. 1". If 
this figure is the projection of a plane quadrilateral ABCD, 
index the plan d. 

2. Let the index of d be 40. (a) Find the angle between the two 

planes of which a 1Q b i0 , c.J>. m are the scales of slope, (b) Find 
the angle between the planes ABC, DEC oi the gauche quadri- 
lateral ABCD. 

3. Determine the scale of the slope of the plane which bisects the 

angle between the planes 10 /' 40 , c : b m of Ex. 2. Also work this 
problem for the planes ABC, DEC of the same example. 



286 PRACTICAL SOLID GEOMETRY chap. 

248. Problem. The indexed plans a 10 , b., of two points 

A, B, and the scale of slope c d., of a plane are given ; it 
is required to determine a point P in the given plane 
which shall be distant 9 units from A and 12 units from 

B. Unit = -05 ". 

Since the required point P is io units from A it must 
be situated on the surface of a sphere with A as centre and 
radius io units. Similarly P must be situated on the 
surface of a second sphere, with B as centre and 1 2 units 
as radius. It is also contained by the given plane. 

Hence the required point is either of the two points 
where the circles in which the plane intersects the spheres 
cut one another. These points are determined in the 
following manner : 

Take cd as a ground line and project cd' the edge 
elevation of the given plane, and also d, b' the elevations 
of A and B. With d as centre and radius 9 units describe 
a circle intersecting cd' in u and r. With b' as centre 
and radius 1 2 units describe another circle intersecting cd 
in w and /. These circles are the elevations of the 
spheres referred to above, and ur\ w's are the elevations 
of the circles in which the given plane intersects these 
spheres. 

Draw dm and b'ri at right angles to cd', then M and N 
are the centres of the circles. 

These circles intersect each other in two points (P and 
<2), each of which satisfies the required conditions ; the 
projections of P will now be determined. 

Conceive the plane with the two circles to be turned 
into the ground about its horizontal trace. We thus 
obtain the rabatments of the two circles, viz. the circles 
with centres M and JV ; these intersect each other in 
two points, one of which is P . Obtain p' by a process 
the reverse of that by which M and JV were obtained 
from m' and ri ; from p' project the plan p n , the index 
being obtained by measuring the height of /'. Thus 
the figured plan of the required point P is found. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 287 




uy / 




! 




0y< -'' ; 




.2*7 




r i : i r i 


Mil! 




^ 




,-'' 1 \ 








La 

! \ \ i V 


^ 
'.-'<? 






a m M i . 
\ " lb l 


/ | 






\2f 


/z< 




A, 



A 7 ^. This problem is of considerable importance in so far that 
many other problems may be reduced to it. 

The point P was determined by regarding it as one of the two 
points of intersection of three surfaces two spheres and a plane. Now 
there are two methods of determining such a point. We may deter- 
mine the intersection with each other of any two of the surfaces, and 
then obtain the points in which this intersection meets the third 
surface ; or we may determine the intersections of one of the surfaces 
with each of the other two, and afterwards obtain the points in which 
these two intersections meet each other. The latter method has been 
adopted in the above solution. 

* 

Examples. 1. Draw a quadrilateral (/';so f 28 </ 'i5 ma ^ n g rt (/';so = 2 > 
b vf<8L = 1 5 Vis = 7 5 <V-28 = 2 > ^15 = J 7- Regard a^ as 
the scale of slope of a plane, and determine the indexed 
plan of a point on the plane, distant 10 units from C, and 15 
units from D. Unit = 0.1". 

2. In Ex. 1 determine the indexed plan of a line lying in the given 
plane and making (>o with the given line. 



2S8 PRACTICAL SOLID GEOMETRY cuap. 

249. Problem. To determine the shortest line MM 
between two given lines, AB, CD. 

Let a i 1> V c\ d Q be the given indexed plans of the lines 
AB, CD. 

Suppose that from any point in one of the lines, say B 
in AB, a line BE be drawn parallel to the other CD. 
Then ABB determines a plane through AB parallel to 
CD. Let now CD be projected on this plane, and let the 
projection cut AB in M. At M erect a perpendicular to 
the plane. This perpendicular will meet CD in the point 
we have called N. Then MN\% the shortest line required. 
It is perpendicular to both AB and CD. 

Through b 12 draw b Y < % parallel and equal to c 10 d , the 
index 2 of e being determined so that the difference of the 
indices of b and e is the same as that of c and d, viz. 10. 
Then BE is parallel to CD. 

Determine^ and g , the traces of BA and BE; or by 
Prob. 240 determine the plan of any other horizontal line 
in the plane of ABE. 

Draw xy perpendicular to the horizontal line ; and on 
xy project the elevations f, a', b', c , and d'. Then fa'b' 
is the edge elevation of the plane ABE, and c'd' will be 
parallel to fab'. 

Select any point in c'd', say c , and draw c'h' perpendicular 
to a'b'. Project from }i to //, where c lQ h is perpendicular 
to f g . Through h draw km parallel to d e 1() . Draw mn 
perpendicular X.0 f^g^ 

Then MN is the shortest line required. The indices 
of m and n may be found by projecting tri, n as shown, and 
then measuring the heights of these points above xy. 

Examples. 1. Two lines </' 18 , c^t^, each 2.5" long, which 
bisect one another at 6o, are the indexed plans of two lines 
AB, CD. Determine the indexed plan of MN, the shortest 
line between them. 

2. Measure the angle between AB and CD. 

3. Determine a line PQ which meets AB and CD each at an angle 

of 6^. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 289 




P2' 



4. Draw a triangle aoc, having ac 



2i" 



ao = 1 



3" 



-ll' 



Regard 



ac as the plan of one edge of a regular tetrahedron ; the plan 
of an adjacent edge AB coincides, in direction only, with ao ; 
complete the plan of the tetrahedron, the indices of a and c 
being 12 and 25 respectively. Unit = o. 1". 

Hint. See Prob. 248. Find the true length of an edge AC 
of the tetrahedron ; then the point B is fixed in the following 
way. (1) It lies on a sphere, centre A, radius AC; (2) it lies 
on a sphere, centre C, radius AC; (3) it lies on a vertical 
plane whose plan is given by the line ao. Therefore draw the 
projections of these spheres, and find (by rabatting the vertical 
plane) the circular sections of the spheres by the vertical 
plane ; the circles intersect in two points either of which is 
the rabatment of B. The plan b may then be readily 
determined. 

A line a () l>., n , 2 2 " ' on S> ^ s tne indexed plan of a square ABCD ; 
unit = o. 1". The plan ad of an adjacent side makes 45 with 
ab. Complete the indexed plan of the square. 

Suppose ABCD of Ex. 5 to be the base of a right pyramid 3" 
long, vertex V; draw the indexed plan of the pyramid. 

Draw the indexed plan of an equilateral triangle ABC, having 
given a d 1Q = 2", a i\ Q c=s\ 

U 



290 PRACTICAL SOLID GEOMETRY chap. 

250. Miscellaneous Examples. 

*1. Determine a plane, bisecting the angle between the two given 
planes. (1886) 

*2. ab is a given line, CD the scale of slope of a given plane. 
Determine the projection of the line ab on that plane. 
Unit = o.i". (1897) 

*3. Determine the intersection of the three given planes A, B, and 
cod. Unit 0.1 ". (1S88) 

*4. cab is the horizontal trace of a plane inclined at 40 to the 
horizontal plane, cd is the plan of a line CD in that inclined 
plane. Draw the traces of a plane inclined to the first plane 
at 40 , the intersection of the two planes to be the line 
CD. (1897) 

*5. The figured plan of a triangle ABC is given ; ef is the plan of a 
line which is bisected by the plane of the triangle ABC. 
Obtain the index of/. (1S93) 

*6. Find the common perpendicular to the two lines AB, CD, and 
give its length. Unit = o.i". (1897) 

*7. Two non-intersecting lines AB and CD are given (see figure 
for Ex. 6). Determine the traces of a plane containing CD, 
and parallel to AB, and determine the projection of AB on 
this plane. 
8. Two lines ab, cd, at right angles, are the plans of the centre 
lines of two horizontal shafts, AB, CD, one of which is two 
feet above the other. They are connected by a third shaft, 
intersecting them in P and Q, such that PQ makes angles of 
60 and 45 respectively with AB and CD. Draw the plan of 
PQ. Scale ^ inch to the foot. 

*9. Determine the angle between AB (Ex. 6) and a line AE which 
intersects CD and lies in the same vertical plane as AB. 
*10. AB and CD (take the figure of Ex. 6, but alter the index of b 
to 16), two non-intersecting straight lines, are given by their 
figured plans. Frflm A draw a line making an angle of 45 
with AB and intersecting CD. Unit o. 1". 

Hint. The required line must lie on the surface of a cone, 
vertex A, axis AB, semi-vertical angle 45 ; it must also be in 
the plane A CD. Draw the plan of the cone, and take any 
vertical plane perpendicular to AB cutting the cone in a circle ; 
draw the elevation (on this vertical plane) of the circle. Take 
any two points on CD, say C and D. Find the points E and F 
in which AC and AD meet the vertical plane, then either of 
the points, in which the line EF meets the circle, when joined 
to A will give a line satisfying the required conditions. 

Observe that EF is the intersection of the plane A CD and 
the vertical plane cutting the cone. 



xi HORIZONTAL PROJECTION, OR FIGURED PLANS 291 




CHAPTER XII 

PLANE AND SOLID FIGURES IN GIVEN POSITIONS 

251. How position is defined. Let the student take a 
square cut out in paper, and place it so as to lie on a plane 
surface which is inclined at a given angle, one side of the 
square making another given angle with horizontal trace of 
the plane ; then he will observe that the shape of the plan is 
always the same no matter what position the square may 
occupy on the plane, so long as the two angles remain 
unaltered. The shape of the plan in this case is therefore 
made definite by two angles being given. If we suppose 
the square to be one face of a cube, then the shape of the 
plan of the cube is also definite. 

The angular position relatively to the horizontal plane 
may be fixed in other ways ; for example, by having given 
the inclination of two sides ; or, what amounts to the same 
thing, the differences in the heights of three corners of the 
square. The shape of the plan is again definite. 

The shape of the elevation is not completely defined by 
the conditions just stated, but depends further on the 
position which the object occupies relatively to the vertical 
plane of projection. If the shape of the elevation is to 
be made definite as well as that of the plan, one more 
angle must be given ; e.g. the angle which one side makes 
with the vertical plane. 



chap, xii FIGURES IN GIVEN POSITIONS 293 

Reasoning thus, it is seen that the shape of the plan of a 
polyhedron is definite when any one of the following sets 
of conditions are given : 

(a) The inclination of a face and that of a line in the 

plane of the face. 

(b) The inclinations of two lines, or the heights of two 

points above a third, all connected with the solid, 
if) The inclinations of two faces. 

(d) The inclination of a face, and that of a line con- 

nected with the solid, not in the plane of the face. 
The shape of the elevation is also definite if, in addition, 
one of the following conditions be given : 

(e) The inclination of a line to the vertical plane, or the 

difference of the distances of two points from the 
vertical plane. 
(/) 77ie inclination of a face of the solid to the vertical 

plane. 
It should be observed that the conditions stated in the 
preceding article do not define completely the position 
of a figure in space relatively to the surrounding objects, 
but only its angular position. Complete definition of 
position may be obtained as follows: 

Let three planes mutually perpendicular be taken as 
planes of reference, then the position of a point in space is 
defined if we know 

(g) The distance of the point from each of the three 

planes of reference. 
The position of a finite line is definite if we know 
(//) The position of one point as in (g), and the inclina- 
tions of the lifte to two of the planes of projection. 
The position of a polygon or polyhedron in space is 
defined if we know 

(/) The position of one point as in (g) and also the 
angular position relatively to the planes of refer- 
ence. To define the latter three angles are necessary, 
which may be those of (a) and (e) in the preceding 
article, or other combinations. 



294 PRACTICAL SOLID GEOMETRY chap. 

252. Problem. To draw the plan of an equilateral 
triangle, l 1 " side, the plane of which is inclined at e, and 
one side at a. 

Draw a plane inclined at 0, and in it draw any line AB 
inclined at a, and obtain bA the rabatment of the line as 
described in Prob. 191 and here repeated. 

Draw an equilateral triangle C D Q E of the given size, 
with one side parallel to, or in, bA ; let this be the rabat- 
ment of the triangle about th. The plane of rabatment is 
now to be turned back into its original position vth, carrying 
the triangle along with it ; the projections of the triangle 
will then be cde, c'd'e', obtained by the construction of 
Prob. 186 reversed. The circular arcs with / as centre are 
the elevations of the paths of C, D, E, the plans of these 
paths being C c, -D d, E^e, perpendicular to ///. 

The problem is thus solved. The plane of the triangle 
is inclined at 6, and the side CD at a to the ground. 

253. Problem. Draw the plan of a cube of given 
edge, with one face inclined at e, and a diagonal of that 
face at a. 

Begin, as in the last problem, by finding B M , the 
rabatment of a line inclined at a, lying in a plane inclined 
at 6. The construction lines for this are not shown. 

Draw the square A B Q C Z> , of the given side, with the 
diagonal A C parallel to E M . Then A Q B C D may be 
taken as the rabatment of the face of the cube, and also as 
the plan of the cube in the rabatted position. 

The plane of rabatment must now be turned back into 
the original position vth, and the elevation of the face 
ABCD in the plane obtained as in the last problem. 

The elevation of the cube is completed by drawing the 
lines a'd , b'b', c'c' , d'd' perpendicular to tv, each equal to the 
edge of the cube, and then joining a'c. 

The plan of the cube is now found by projecting the 
points in elevation on to the lines through corresponding 
points of the rabatment, drawn at right angles to th. 



XII 



riC.URES IN GIVEN POSITIONS 



295 









Examples. 1. Draw the plan of a square, .2" side, the plane of 

which is inclined at 50 , one side being inclined at 35. 
2. Draw the plan of a square ABCD, 2" side, when the line joining 
A to the middle point of BC is inclined at 35 and the plane of 
the square at 50 , the point A being on the ground. 

Determine the inclinations of the diagonals of the square. 

Draw the plan of a regular hexagon of ij" side in any position 
such that its plane is neither horizontal nor vertical. 

A regular hexagon of ij" side has one side in the horizontal 
plane. The plane of the hexagon is vertical, and inclined at 
43 to the vertical plane of projection. Draw the elevation of 
the hexagon. 

Draw the plan of a cube of 2-|" edge, one face being inclined 
at 50 and one side of that face at 35. 

An isosceles triangle, base 2.5", sides 3", has its base inclined at 
35, and its plane at 50 ; this is the end of a right prism 2^" 
long. Draw the plan and elevation of the solid. 

Draw the plan of a square pyramid, side of base 2", height 3", 
when the base rests on a plane inclined at 60, one diagonal of 
the base making 50 with the horizontal trace of the plane of 
the base. 



3. 

4. 

5. 

6. 



296 PRACTICAL SOLID GEOMETRY chap. 

254. Problem. An octahedron, 2" edge, has one face 
resting on a plane inclined at 30 , one side of the face 
making an angle of 70 with the horizontal trace of the 
plane. Draw the plan and an elevation of the solid. 

Let VTHhe. the plane inclined at 30 . Draw any line 
Z 7l/ making an angle of 70 with the horizontal trace, and 
construct an equilateral triangle A B Q C , having one side in 
the line Z Af Q . Let A B C be the rabatment of that face 
ABC of the octahedron which rests on the plane at 30 , 
AB being the side of the face which makes an angle of 70 
with the horizontal trace. Complete the plan of the solid 
in its rabatted position as follows : 

Inscribe a circle in A B C , and draw f x d v d x e v e x f v 
tangential to the circle, respectively parallel to A B , C Q A , 
B C . Then d x e x f x is an equilateral triangle, and is the 
plan of the upper face of the octahedron. The plan of the 
solid is completed by drawing the outline shown. 

Now let the plane of rabatment be turned back to the 
original position vth ; the elevation a'b'c of the face ABC 
is readily found as in previous problems, and the construc- 
tion is evident from the figure. . 

To determine the elevation of the face DEF, first find 
the distance between the parallel faces ABC, DEF. This 
is done in the figure by finding B B> , the rabatment of 
BD about its plan B d y That is, d x D is drawn at right 
angles to the plan, and B D made equal to BD, \\" . 
Then d x D is the distance between the faces. 

To find the elevation of D determine the point s, as 
shown, / being the elevation of the foot of the perpendicular 
from D to the plane of the face ABC, and draw s'd' per- 
pendicular to tv and equal to d x D . In a similar manner 
e and/" may be found. The elevation of the octahedron 
is then completed by drawing the lines representing its 
edges. 

The determination of the plan is left as an exercise. 

Example. An octahedron 2" edge has a face inclined at 6o, 
and one side of that face at 40'. Draw its plan. 



XII 



FIGURES IN GIVEN POSITIONS 



297 




Examples on Problems 255 to 258. 

1. The sides of a square are inclined respectively at 30 and 45 ; 

draw the plan and determine the inclinations of the diagonals. 

2. Draw plan of a square pyramid, side of base I7?", height 1^", 

when two sides of the base are inclined at 30 and 40 re- 
spectively. Determine the inclinations of the sloping edges. 

3. Determine the plan of a hexagon of 1" side, when two adjacent 

sides are inclined 30 and 40 J respectively. 

4. Draw the plan of a regular hexagon ABCDEF, \\" side : (a) 

when two diagonals are inclined at 30 and 5 respect- 
ively ; (l>) when two alternate sides are inclined at 30 and 
50 respectively ; (r) when AE and CF are inclined at 30 and 
50 respectively. 

5. Draw the plan of a regular tetrahedron, 2" edge, when a face 

and an edge are inclined respectively at 55 and 40. Show 
an elevation of the solid on a vertical plane which makes an 
angle of 30 with the edge which is inclined at 40 . 

6. Obtain the projections of an equilateral triangle, 2" side, when 

two sides are inclined at 30 and 50 to the horizontal plane, 
and the third side at 40 to the vertical plane. 

7. Two lines meet at an angle of 60. The plane containing them 

is inclined at 40 , and one of the lines is inclined at 50. Find 
the inclination of the other. ' 



298 PRACTICAL SOLID GEOMETRY chap. 

255. Problem. To draw the plan of a square, having 
given the inclinations a t and a., of two of its sides. 

Let AB, AD be the sides of the square which are in- 
clined respectively at a x and a. 2 , and suppose the point A 
to be on the ground. 

Commence the solution by drawing a square A B Q C D ; 
let this be the rabatment of ABCD from the required 
position, into the ground, about the horizontal trace of the 
plane containing the square. 

The horizontal trace must now be found ; it is a line 
through A such that if the rabatted square be turned back 
about this trace, until AB be at the inclination a v then at 
the same time AD shall be at the inclination a 2 . 

Draw the right-angled triangles A 1 B x b v A x D x d v having 
the base angles respectively equal to a x and a. and the 
sides A X B V A X D X each equal to the side of the square ; 
draw D X H X parallel to the base. Then, as in Art. 179, 
B x b x , D x d x are the heights of B and D above A, i.e. above 
the ground ; and A x b v A x d x are the lengths of the plans of 
AB, AD. Also H x in A l B 1 determines the position of a 
point Zfin AB, 'which is at the same height as D. 

Make A Q If Q = A X H V and join D Q H . Then D If is 
the rabatment of a horizontal line in the plane of the square. 
The horizontal trace of the plane of rabatment is therefore 
a line through A parallel to D If ; it is denoted by th in 
the figure. 

Finally, the required plan may be obtained in two ways. 

First Method. With centre A () , radii A x b x , A x d x , the 
lengths of the plans of AB, AD, describe arcs intersecting 
the lines dawn through B , D Q , perpendicular to th, in b 
and d respectively ; join ab, ad, and draw dc, be parallel to 
ab, ad. Then abed is the required plan of the square. 
B b, D d are the plans of the circular paths of B and D 
traced during the rabatment. 

Second Method. Take an xy perpendicular to D Q II Q , or 
/// ; draw the two horizontal lines mm, nn, at heights above 
xy equal to B x b v D x d^ respectively. With centre /, describe 



XII 



FIGURES IN GIVEN POSITIONS 



299 




arcs through B^,D^ (projected from fi , D { ), intersecting the 
horizontal lines mm, nn in // and d' respectively. Then //, d' 
are the elevations of B, D, a' is the elevation of A, and a, 
d', b' will lie in one straight line, which is the edge eleva- 
tion of the plane of the square. The arc C Q 'c determines 
c ; and the plan of the square is obtained by drawing pro- 
jectors from the points in elevation, to intersect lines from 
the corresponding points of the rabatment, drawn at right 
angles to th, the horizontal axis of rotation. 

A model may be made to illustrate this problem. 

Draw a square ABCD, Fig. ((7), and draw the lines Ab, Ad, making 
angles of a v a 9 with AB, AD. Draw Bb, Dd perpendicular to Ab, 
Ad. Make bo equal to Dd, and draw oH parallel to bA. Cut out 
this figure in paper, then indent and fold the triangles ABb, ADd 
about AB, AD to such a position that when Ab, Ad rest on the ground, 
b and d are the plans of B and D. 

By studying this model, the reasons for drawing the 
various construction lines should be quite evident. 



300 PRACTICAL SOLID GEOMETRY chap. 

256. Problem. To draw the plan of a regular hexagon 
of given side : (a) when the inclinations of two diagonals 
are given ; (b) when the inclinations of two alternate sides 
are given. 

Draw A B Q C D E (j F , the rabatment of the hexagon. 

(a) The rabatment C Ff Q of a horizontal line is found 
with the aid of Fig. (a), as in Prob. 255. The horizontal 
trace of the plane of rabatment is parallel to F Q H , and 
may be taken through Q . 

The plan is then determined as in Prob. 255, either 
by the first or second method. If the first method be 
adopted, and the plan of OAF be thus found, the plan of 
the hexagon can be completed, without drawing an eleva- 
tion, by making use of the following proposition of pure 
solid geometry : "If a number of straight lines are parallel 
to each other and of equal lengths, then their projections 
on any plane are also parallel to each other and of equal 
lengths P 

Further, the elevation on any xy can be obtained with- 
out first drawing the edge elevation, by making use of the 
properties of the perpendiculars A x a v B x b v as explained in 
Art. 179. 

(b) Let the sides AF, DE be those having the given 
inclinations. Produce them to meet in P. 

Then by the aid of Fig. {I?) the rabatment F> A' of a 
horizontal line in the plane of the hexagon is determined. 

The horizontal trace of the plane of rabatment is parallel 
to DqKq, and may be taken through F Q . 

The plan can be obtained by either of the methods of 
Prob. 255, making use of the proposition stated above if 
the first method be the one adopted. 

257 Problem. An equilateral triangle has two of its 
sides AB, AC inclined at a x and a 2 to the horizontal plane, 
and the third side BC inclined at 6 to the vertical plane ; 
determine its plan and elevation. 

Obtain the plan as in Prob. 255, by first drawing the 



xir 



FIGURES IN GIVEN POSITIONS 



301 




rabatment A B Q C ; then by the aid of the triangles A l B l b v 
^\C\ C \ f m d C H the rabatment of a horizontal line; and 
finally determine the plan abc by the first method. 

To find the elevation, first draw a right-angled triangle 
B 2 C/,, where B 2 C. 2 = BC, and the angle C 2 B 2 c 2 = ft. With 
centre c, radius C\/ 2 , the difference in the distances of B and 
C from the vertical plane (Art. 179), describe the circle 
shown, and draw the tangent bm. 

Then the plan abc and an elevation on an xy parallel to 
bm will satisfy the conditions. Compare with Prob. 1S0. 



302 PRACTICAL SOLID GEOMETRY chap. 

258. Problem. Two lines meet at an angle of 65 '. The 
plane containing them is inclined at 50 , and one of the 
lines is inclined at 40 ; find the inclination of the other. 

Represent a plane inclined at 50, in which place a line 
AB inclined at 40 , and find its rabatment A B . 

Draw a line A C making 65 with A B . 

Determine the plan and elevation of AC when the plane 
of rabatment has returned to its original position. 

Then find the true inclination of AC. 

259. Problem. Draw the plan of a regular tetrahedron, 
\V edge, three of its corners heing at heights of -]-", \\", 
and 11" respectively ahove the ground. 

Draw A Q B Q C Q d x the plan of the tetrahedron, with the 
face ABC on the ground, and consider this the plan of the 
solid in its rabatted position. 

In the two triangles shown to the right, make A 1 B 1 = 
A l C 1 = 1 1", the length of edge of the solid, and B x b v C x c x 
respectively equal to 1-}" and 1", the heights of B and C 
above A. Draw C X H X parallel to c x A v and make A Jd = 
A X H X ; then C Q J7 is the rabatment of a horizontal line in 
the face ABC. 

Take xy perpendicular to C Q Jif Q , and determine a'b'c the 
elevation of ABC, as in Prob. 255, second method, assum- 
ing the point A in the ground, and drawing the horizon- 
tal lines nun, /in, at distances equal to B x b x , C x c x , above xy. 

To determine the elevation of D, first find d x B> () , the 
distance of D from the face ABC, by the rabatment of 
CD about its plan as in the figure, or by any other method. 
Obtain the point s as shown, and draw s'd' perpendicular 
to db' and equal to d x B> Q . Then d' is the elevation of D, 
and the elevation of the tetrahedron is completed by draw- 
ing the lines representing the edges. 

The plan is to be obtained as in previous problems, and 
is left as an exercise for the student. 

Finally, if an xy be drawn parallel to the one shown, and 
|" below it, all the conditions of the problem are satisfied. 



XI f 



FIGURES IN GIVEN rOSITl<>\- 



.-301 



H./-\~tC, 




b. c> 



Examples. 1. Determine the plan of a hexagon of f " side, when 
three alternate angular points are i", i|", and if" high re- 
spectively. 

2. Draw the plan of a square 2^" side, when the heights of its 

centre and two corners, not opposite each other, are i", if", 
and 2V respectively. 

3. Draw the plan of an isosceles triangle, base 2^", sides 2", when 

the extremities of the base and the middle point of one of the 
sides are at heights of f", 1^", and 2" respectively. 

4. Draw the plan of an equilateral triangle, 3" side, when the 

three middle points of the sides are at heights of 1", l%", if" 
respectively. 

5. Draw the plan of a regular tetrahedron, 2\" edge, the heights of 

three of its corners above the horizontal plane being respectively 

2 ' *~2 ' ^ 

6. Draw the plan of a cube, 2^" edge, when three of its angular 

points are at heights of 1", 1.25", and 2" above the horizontal 
plane. Make an elevation on a plane parallel to one of the 
diagonals of the solid. 

7. An isosceles triangle is the plan of an equilateral triangle. Find 

the inclination of the plane of the triangle (1) when the equal 
sides are each three-fourths of the base ; (2) when the base is 
three-fourths of each of the equal sides. 

8. Draw the plan of an octahedron of 2" edge, when two diagonals 

of the solid are inclined at 26 and 36 respectively. 

9. Three corners of a square, 2" side, are 1", 1.4", and 2.1" high ; 

find the height of the centre. 



304 PRACTICAL SOLID GEOMETRY chap. 

260. Problem. To draw the plan of a cube having 
given the length of edge and the inclinations e and 9 of 
two faces. 

Draw the traces vth of a plane inclined at 0. 

By Prob. 233 determine a plane at right angles to VTH 
and inclined at <f>. That is, choose any point A in the . 
plane and draw the projections of a cone, vertex A, base 
on ground, base angle <. Find also the horizontal trace 
JV of a line through A perpendicular to the plane VTH 

Then the tangent nlm is the horizontal trace of the 
plane at <. 

The line ias is the plan of the intersection of the planes. 
If now one edge of the cube coincide with I A, and two 
faces with the two planes, the conditions will be satisfied. 

Rabat the plane VTH, and thus obtain t'S . 

Draw the square with one side in t'S ; this is the plan 
of the cube while on the ground. Now let the plane VTH 
be raised to its proper position, and draw the corresponding 
plan and elevation of the cube. 

261. Problem. To draw the plan of a tetrahedron, 
having given the length of edge and the inclinations e 
and 9 of two faces. 

Draw the traces vth of a plane inclined at 6. 

Then by Prob. 234 determine the traces of a plane 
which is inclined at <, and makes with the plane VTH an 
angle a equal to that between the base and one of the equal 
faces of the tetrahedron. 

Determine the intersection of these two planes and 
rabat it, along with the inclined plane, into the ground. 

Proceed now exactly as in the last problem, by drawing 
an equilateral triangle having one side (equal to the given 
edge) in the rabatment of the intersection. 

While the inclined plane is in the ground complete the 
plan of the tetrahedron. Finally suppose the inclined 
plane to be raised to its proper position, and thus complete 
the required plan and elevation of the tetrahedron. 



xn 



FIGURES IN GIVEN POSITIONS 



3o5 




X-A 



\\ /--' V. 







/, 



306 PRACTICAL SOLID GEOMETRY chai\ 

262. Problem. To draw the plan of a cube, having 
given the length of edge, the inclination e of one face, and 
the inclination 9 of a diagonal of the solid. (No figure.) 

There are three preliminary problems to be worked. 

1. Find the length of the diagonal. 

2. Find the angle a between a diagonal and a face. 

3. Find the angle /3 between a diagonal and an edge. 
Having solved these, draw the projections of a diagonal 

inclined at <f>. Since a plan only is required, the diagonal 
may be parallel to the vertical plane. 

Then by Prob. 235 determine a plane inclined at 6, and 
making an angle a with the diagonal. 

Next obtain the plan of an edge which meets the dia- 
gonal and lies in the plane. This is done by Prob. 236, the 
edge making j3 with the diagonal. 

Now rabat the plane with the edge into the ground. 

On the rabatment of the edge construct a square. This 
will be the plan of the cube while one of its faces is on the 
ground. 

Now turn the plane back into its original position, carry- 
ing the cube with it, and thus complete the plan of the 
solid by well-known methods. 

Examples. 1. Draw the plan of a cube, 2^" edge, when two of 
its adjacent faces are inclined at 45 and 75 respectively. 

2. A building brick 9" x 4^" x 3" has one end inclined at 40, and 

a long face at 6o. Draw its plan. Scale \. 

3. A tetrahedron, 2^" edge, has two of its faces inclined at 40 

and 70. Draw the plan and elevation. 

4. A square pyramid, base 2" side, axis 3" long, has its base in- 

clined 40 , and one of its faces at 6o. Draw its plan, and a 
sectional elevation on a vertical plane which bisects any two of 
its long edges. 

5. A cube, 2\" edge, has a diagonal inclined at 6o, and a face at 

65 ; determine its plan. 

6. Draw the plan of a regular tetrahedron of 2" edge, when the 

line joining a vertex to the centre of the opposite face is hori- 
zontal, one of the other faces being inclined at 6o. What is 
the least angle which may be substituted for the 6o, so as not 
to make the solution impossible ? Ans. \0)\. 



xii FIGURES IN GIVEN POSITIONS 307 

263. Miscellaneous Examples. 

1. Draw the plan of a hexagon of 1.5" side, the heights of three 

successive adjacent corners to be 1", 1.5", and 0.75". (1878) 

2. An isosceles triangle (sides z\" , base i|") has its base in the 

horizontal plane and one side in the vertical plane. The base 
makes an angle of 35 with xy. Determine the plan and eleva- 
tion of the triangle. ( J 893) 

3. A square of 3" side lies on a plane inclined at 50 . One side of 

the square makes 40 with the horizontal trace of the plane. 
Draw its plan. (1885) 

4. Draw the plan of a cube of i T V' edge, one face inclined at 50 

and a second face at 60 . (1890) 

5. Draw the plan of an octahedron of 2" edge, one edge being in- 

clined at 30 , and another (on the same face) at 20 . (18S2) 

6. Draw the complete plan of a cube of 1.5" edge, two faces in- 

clined respectively at Co Q and 70 to the horizontal plane. 

(iS95) 

7. A right pyramid has for its base a regular pentagon of which the 

diagonals measure 2.5". The vertex is 2" above the base. 
Draw the plan and elevation of the pyramid, with its base in a 
plane inclined at 55 to the vertical plane, and at 6o to the 
horizontal plane ; one diagonal inclined at 30 , and one end of 
that diagonal in the vertical plane. (1894) 

8. Draw a plane inclined at 45 to the horizontal plane, and at 6o 

to the vertical plane. Draw a line in the plane inclined at 30 , 
and 2" long between its traces. Lastly, draw the plan of a 
regular hexagon lying in the plane of which the above line is a 
diagonal. O897) 

9. An octahedron of 2j" edge has the plane containing two of its 

diagonals inclined at 30, and that containing one of these, and 
the other diagonal inclined at 70 . Draw its plan. 

10. Determine the projections of a cube on three planes mutually 

perpendicular, having given the inclinations of three adjacent 
edges, one to each plane. 

11. Determine the projections of an equilateral triangle on three 

planes mutually perpendicular, having given the inclinations of 
its sides, one to each plane. 

12. Draw the projections of a cube on three planes mutually per- 

pendicular, having given the inclinations of three adjacent faces, 
one to each plane. 



CHAPTER XIII 

THE PROJECTION OF CURVES AND CURVED SURFACES 

264. General method of procedure.- In the problems 
hitherto considered, the figures represented in projection 
have been made up of points, straight lines, and planes. 
The projection of curves, and of solids bounded by curved 
surfaces, is now to be considered. 

The general method of projecting a curve is to first 
find the projections of a number of isolated points in it ; 
then to draw a curve by freehand through the points thus 
determined. 

In projecting a solid, the projections of the straight or 
curved edges are first found ; then if necessary the outline 
of the projection is completed, as determined by projectors 
which touch the surface of the solid. This will be more 
particularly explained in the problems which follow. 

265. Problem. Draw the plan of a circle, \\" diameter, 
when its plane is inclined at 50\ 

First Method. Begin with the rabatment, a circle 
A B , i\" diameter. Take twelve equidistant points on 
the circumference of this circle, as shown. Find the plans 
and elevations of these points when the plane of rabatment 
is turned back into its original position 7'th, inclined at 50. 

The line ab' is the edge elevation of the circle ; and a 
fair curve carefully drawn through the plans of the points is 
the plan of the circle. The plan is an ellipse. 



CH. XIII 



CURVES AND CURVED SURFACES 



309 




o 



Second Method. Begin by drawing an edge eleva- 
tion of the circle, the line a'b', 1^" long, inclined at 50'. 
On a'b' draw a semicircle, which divide into six equal 
arcs as shown. From the points of division draw lines 
perpendicular to a'b', intersecting the latter in o', 1', 2 '. 

Conceive o', 1', 2 as the elevations of chords of the 
circle perpendicular to the vertical plane. Then the dotted 
perpendiculars represent the halves of these chords, when 
half the circle is turned about AB into a position parallel to 
the vertical plane. 

To obtain the plan of the circle, draw projectors from 
the points in a'b'. Consider the projector from 1 ' : let m 
be its intersection with ab ; make mi, m\ each equal to 
i'ij^; repeat this construction for the other points. In this 
way the points o, 1, 2 on the plan of the circle are found, 
and the ellipse is drawn through them. 



310 PRACTICAL SOLID GEOMETRY chap. 

266. Problem. A given sphere resting on the ground 
is cut by a given vertical plane. Draw the plan and 
elevation of the trace of the section plane on the surface, 
and show a sectional elevation of the sphere. 

The projections of the sphere are circles, equal in 
diameter to the sphere, drawn with centres c, c, the plan 
and elevation of the centre of the sphere. Since the 
sphere rests on the ground, the elevation touches xy. 

Let VTH be the given vertical section plane. The 
section is a circle of which ab is the plan. The elevation 
may be found as follows : 

On ab as diameter draw a semicircle, and divide its 
circumference into six equal arcs as shown. From the 
points of division draw lines perpendicular to ab, intersect- 
ing the latter in o, i, 2. 

Let o, 1, 2 be the plans of vertical chords of the circle. 
Then the dotted perpendiculars represent the halves of 
these chords, when half the circle is turned about AB into 
a horizontal position. 

Draw projectors from the points on ab. Consider the 
projector from 2 : let ;;/ be its intersection with the hori- 
zontal line through c ; make 1112, 1112 each equal to 2 Q 2 ; 
repeat this construction for the other points, and through 
the points in elevation thus found draw a curve, which is 
an ellipse, and is the elevation of the required trace of 
the section plane on the surface of the sphere. 

The sectional elevation is that on xy\ taken parallel to 
ab. The method of obtaining this needs no further ex- 
planation. 



Examples. 1. Draw the plan of a circle 2^" diameter, when its 
plane is inclined at 50 . 

2. Draw the elevation of a circle 2\" diameter whose plane is 

vertical and inclined at 60 to the vertical plane. 

3. A hemisphere, 2^" diameter, rests with its flat face on the ground. 

It is cut by a vertical plane h" distant from its centre, which 
makes an angle of 50 with xy. Draw the elevation, showing 
the curve of intersection. 



XIII 



CURVES AND CURVED SURFACES 



3ii 




6. 



A sphere 3" diameter is cut into four equal parts by two 
perpendicular planes through its centre. Draw the plan of 
one of these parts when resting with a flat face on the ground ; 
and give a sectional elevation on a vertical plane which makes 
45' with the straight edge of the solid, and is distant 1" 
from the centre. 

A sphere, 4" diameter, is cut into eight equal parts by three planes 
mutually perpendicular. Draw the plan of one of these parts 
when resting on a flat face. Also draw an elevation on a 
vertical plane, which makes an angle of 30 with a horizontal 
edge of the solid, so as to show the flat faces. 

Draw the plan of the solid of Ex. 5 when two of its edges are 
inclined at 30 and 4o\ And add an elevation on a vertical 
plane, the real angle between which and the third edge is 25. 

Draw the plan of the solid of Ex. 5 when its curved surface rests 
on the ground, two faces being inclined each at 70 . 



312 PRACTICAL SOLID GEOMETRY chap. 



267. Problem. A cone, diameter of base 1\", length of 
axis 2 ", rests with its base on the ground, and is cut by a 
plane inclined at 45, bisecting the axis, (a) Draw the 
plan and elevation of the cone, showing the trace of the 
section plane on its surface ; (b) find the true shape of the 
section ; (c) obtain a development of the surface of the 
cone, showing the trace of the cutting plane on the surface. 

(a) The plan of the cone is the circle ab, with s, the 
plan of the veitex, as centre. The elevation is the triangle 
sab. 

To find the section, take twelve equidistant points on 
the circumference of the base, join these to the vertex, and 
draw the plan and elevation of the twelve generating lines 
thus formed. The intersection of the cutting plane with 
these generators gives twelve points on the curve of section 
required. The elevations of the points are in c'd', and the 
plans are found by projecting from the points in elevation. 

A special construction is required to determine the plans 
o, o ; in this case it is evident that so, so are each equal to 
o'o", where do" is parallel to xy. The curve through the 
plans of the points is an ellipse. 

(b) The true shape of the section may be found by a 
rabatment of the cutting plane about its vertical trace, which 
is equivalent to an auxiliary plan on tv. The rabatment 
of CD is c x d v and to obtain the points on the curve, make 
m t i v in, i, each equal to i/n ; and repeat this construction 
for the other points. The true shape of the section is an 
ellipse. 

(c) To obtain the development. With S as centre, 
describe an arc with radius equal to s'd, the length of a 
generator. 

Take any point A on this arc, and by Rankine's con- 
struction, Prob. 113, make arc AT equal to \ circumference 
of base of cone; and set off TB = AT. Subdivide each 
of these arcs into three equal parts. The development of 
the curved surface of the cone, with that of the twelve 
equidistant generators, is thus obtained. 



XIII 



CURVES AND CURVED SURFACES 



3i; 




vvy 



Make SO = so" = true distance of the vertex from the 
point O ; and repeat this construction for the other points. 
A curve through these points is the development of the 
trace of the cutting plane on the surface of the cone. 

Examples. 1. Determine the true shape of the section of a cone 
2h" base, 3" axis, by a plane parallel to its axis, the distance 
between the plane and the axis being j". Develop th cone 
and show the curve of section. 

2. Determine the shape of the section of a cone 2" base, 3" axis, by 

a plane parallel to a tangent plane, ap.d distant J" from the 
latter. 

3. A cone of which the altitude = radius of base= ii" is cut in two 

by a plane containing the vertex, the trace of the plane on the 
base bisecting a radius at right angles. Draw the plan of the 
smaller part when resting with its triangular face on the ground ; 
and obtain an elevation on a ground line which makes an angle 
of 40 3 with the chord of the segmental base. 



314 PRACTICAL SOLID GEOMETRY chap. 

268. Problem. A cone, diameter of base \\", length of 
axis 2", rests with a generator on the ground, the axis 
being parallel to the vertical plane. Draw the plan and 
elevation of the cone ; and add a sectional elevation on a 
vertical plane which bisects the plan of the axis at an 
angle of 45. 

The elevation of the cone is the triangle s'db\ copied 
from the triangle sab' of Fig. 278, but with the generator 
SA on the ground. 

The plan of the circular base is an ellipse determined as 
in Prob. 265, second method; and the plan of the cone 
is completed by drawing the two tangents to the ellipse 
from s. 

Each of these tangents is the line generated by the foot 
of a vertical projector, which projector moves so as always 
to touch the curved surface of the cone. It may be shown 
by pure solid geometry that the moving projector touches 
the surface along two generators, one on each side of the 
cone ; the two tangents from s to the ellipse are the plans 
of these generators. 

Let ////, bisecting sc at 45 , be the plan of the vertical 
section plane. To obtain the sectional elevation, take a 
new ground line x'y parallel to Im, and find the elevation 
on x'y of the vertex and the points on the base according 
to the rules of Art. 170. 

Draw the twelve generating lines in both plan and 
sectional elevation, whence twelve points on the section are 
at once found by projection from the plan. 

The drawing of this view is left as an exercise for the 
student. 



Examples. 1. A cone, 2" base, 3" axis, rests with a generator on 
the ground ; draw its plan, and a sectional elevation on a vertical 
plane which bisects the plan of the axis at 45 . 

2. Draw the plan and elevation of a cone 2" base, 3" axis, when 
the axis is inclined at 30 to the horizontal plane and 45 to 
the vertical plane. Show the section by a horizontal plane 
which bisects the axis. 



XIII 



CURVES AND CURVED SUREACES 



315 




Examples on Problems 270 to 272. 

1. A cylinder, i}/ diameter, 2" long, has a cone 2f" base, i^" 

axis, placed centrally on one of its ends. Draw the plan of 
the solid when the bases of both cone and cylinder touch 
the ground. Draw a sectional plan on a plane which con- 
tains the vertex and two points on the circumference of the 
base of the cone 2^" apart. 

2. Draw a semicircle 5" diameter, in which inscribe the largest 

possible circle. The semicircle is the development of a cone, 
and the circle that of a line on its surface. Draw the plan and 
an elevation of the cone when resting with its base on the 
ground, showing the line on its surface. 

3. Draw a sector of a circle 3" radius, containing an angle of 120 . 

This sector is the development of the surface of a cone. Deter- 
mine the plan and elevation of the cone when resting with its 
base on the ground. 

Suppose a fly, starting from a point in the circumference of 
the base, to walk round the surface and return to the same 
point. Show the plan and elevation of the path, when its 
length is the least possible. 



316 PRACTICAL SOLID GEOMETRY chap. 

269. Problem. A cylinder, 2" diameter, is 3" long. 
Draw the plan and elevation : (a) when the axis is parallel 
to the vertical plane, and inclined at 45 to the horizontal 
plane ; (b) when the axis is inclined to both planes of pro- 
jection. 

(a) Draw the rectangle a'a'b'b' having b'b'=$'\ b'a =2", 
and b'b' making an angle of 45" with xy. Draw c'c bisecting 
the short sides of the rectangle. Then a'a'b'b' is the eleva- 
tion of the cylinder, and c'c that of the axis ; a'b\ a'b' are 
the elevations of the two circular ends. 

The plan is obtained by projecting the two circular ends 
from the elevation, in the manner of Prob. 265, second 
method. The two ellipses thus found are joined by two 
common tangents as shown in the figure, and the plan 
is complete. 

(b) An elevation on any line xy\ not parallel to cc, 
determined according to the principles of Art. 170, will 
satisfy the requirements of (b). The two ends project into 
ellipses, common tangents to which complete the elevation. 

The common tangents to the ellipses, and the lines 
a'a', b'b', are the projections of the generators along which 
the projectors touch the curved surface of the cylinder in 
each case. 

Examples. 1. A cylinder, 2" diameter, is 2|" long, draw the 
plan and elevation : (a) when the axis is parallel to the vertical 
plane, and inclined 40 to the horizontal plane ; (b) when the 
axis is inclined at 50 and 40 to the vertical and horizontal 
planes of projection. 

2. A cylinder, 2V base, 3^" long, is cut into two parts by a plane 

containing the axis. Draw the plan of one of these parts, when 
resting with a rectangular face on the ground, and draw a 
sectional elevation on a vertical plane bisecting the axis at an 
angle of 45 . 

3. A cylindrical cheese is 16" diameter, and 5" thick. A wedge- 

shaped piece is cut out by two planes containing the axis and 
including an angle of 50 . Draw the plan of the piece when 
resting with one of its rectangular faces on the horizontal plane, 
and draw an elevation on a plane making 25 with the short 
sides of that face. Scale ^. 



XIII 



CURVES AND CURVED SURFACES 



317 




270. Problem. To find the projections of the shortest 
line on a given developable surface, connecting two given 
points on the surface. 

First read Prob. 271. The line in question develops 
into a straight line, joining the developments of the points. 
For the length of the line on the surface is equal to the 
length of its development, and when the development is 
straight it is the shortest line possible between the points. 

Therefore, first find a development of the given surface 
and of the given points in it. Draw the straight line between 
the developed points. Then working backwards from 
the development, determine the projections of the line. 
See Fig. 271. 

Note. The shortest line between two points on a cylinder is a helix 
of uniform pitch ; for the development of this curve is a straight line. 
See Probs. 369 and 130. 



318 PRACTICAL SOLID GEOMETRY chap. 

271. Problem. The given sector of a circle SBAB is a 
development of the curved surface of a cone ; and HK 
that of a line on the surface. Draw the plan and an 
elevation of the cone, when resting with its base on the 
ground, showing the line HK. 

The diameter of the base of the cone must first be 
found. Bisect the arc BB in A, and the arc AB in T. 
Draw the tangent AR and make AR equal to the chord of 
\ the arc AT. With centre R describe an arc through T, 
intersecting in / the line At drawn at 45 to SA produced. 
Draw tC perpendicular to SC. Then the circle with centre 
C, radius CA, is the development of the base of the cone. 
For by Rankine's construction, Prob. 113, arc At arc AT 
="| .circumference of base. 

The plan of the cone is thus drawn ; and the altitude 
and the elevation may be found, since the diameter of the 
base, and the length of a generator, viz. SB, are known. 

To obtain the plan of HK, take a series of generators 
intersecting the line. In the figure, four of twelve equi- 
distant generators are shown. The construction is then 
equivalent to that of Prob. 267 (c), worked the reverse 
way, and the completion is left as an exercise. 

272. Problem. The plan and elevation of a solid are 
shown in the figure at (a). Draw the plan when the solid 
is in the position shown in elevation at (b). 

The plan of the cylindrical portion of the solid is found 
as in Prob. 269. 

The curves on the four side faces of the solid are equal 
and similarly placed circular arcs, being the intersections of 
the faces with the spherical surface. The points A, B, C 
on these arcs are similarly and symmetrically taken on the 
four faces. Hence the distance between b, b in plan is the 
same as between b', b' on the front face in elevation. This 
consideration enables the plans of the arcs to be found. 
The plan of the solid is completed by projecting from the 
elevation the edges which are straight. 



XIII 



CURVES AND CURVED SURFACES 



319 





320 PRACTICAL SOLID GEOMETRY chap. 

273. Some properties of the cone and cylinder. Before 
proceeding with the remaining problems of this chapter we 
wish to amplify some of the theorems which are given in 
the Appendix. 

Sphere inscribed in cone or cylinder. Take two inter- 
secting lines of indefinite length and draw the bisector of 
the angle between them with any point on this bisector as 
centre draw the circle touching the lines. Suppose now 
these lines and the circle to rotate about the bisector as 
axis, then it is evident that the lines will generate a cone, 
and the circle a sphere inscribed in the cone, and the sphere 
will touch the cone in a circle perpendicular to the axis. 
It is readily seen that an indefinite number of spheres of 
varying sizes can be inscribed in a cone, all having their 
centres in the axis. 

A cylinder, being the limiting case of a cone where the 
vertex is at an infinite distance away, may also have an 
unlimited number of spheres inscribed in it. All these 
spheres will have a diameter equal to the diameter of the 
cylinder. 

Two cones and a common inscribed sphere. It can be 
shown that if two cones circumscribe the same sphere, their 
surfaces intersect each other in two ellipses. Also, since a 
cylinder is a particular case of a cone, a similar remark 
applies to two cylinders, and to a cylinder and cone. This 
property may be frequently taken advantage of to facilitate 
the working in certain problems. 

Projection of a cone or cylinder of indefinite length. A 
cone of indefinite length may evidently be defined by any 
two spheres inscribed in it, for if these two spheres are 
given, then the line of the axis, the position of the vertex, 
and angle of the cone are known. One of the spheres may 
be at the vertex and so become a point ; thus a cone of 
indefinite length is defined by its vertex and an inscribed 
sphere. The projection of the cone on any plane consists 
of the two tangents drawn to the two circles which are the 
projections of the two inscribed spheres. It must be 



XIII 



CURVES AND CURVED SUREACES 



321 




specified whether the two external or two internal tangents 
are to be taken. As before, one of these circles may be a 
point, i.e. be the vertex of the cone. Similar remarks apply 
to a cylinder of indefinite length, but now the two inscribed 
spheres are of equal size. A cylinder and cone projected 
on the horizontal and vertical planes are illustrated in 
the figure. These indefinite cones and cylinders may be 
drawn broken at the ends as shown. 

These statements must be qualified as follows. If the 
axis of the cylinder be perpendicular to the plane of 
projection, the projection of the cylinder is a circle which 
is the edge view of the surface. 

If the projection of the vertex of the cone fall within the 
projection of an inscribed sphere, the common tangents 
cannot be drawn ; in this case the indefinite cone has no 
definite form of projection, but it may still be represented 
in projection by the circle and point or by two circles. 

Y 



322 PRACTICAL SOLID GEOMETRY chap. 

274. Problem. To determine the horizontal trace of a 
given cone of indefinite length, whose axis EV is parallel 
to the vertical plane. 

The cone may be given by its vertex V, vertical angle a, 
and inclination of axis 6 ; or by its vertex and an inscribed 
sphere. In either case the projections in outline are readily 
drawn ; let them be as shown in the figure. The horizontal 
trace is a conic section, in the present case an ellipse. 

To find the major axis of the ellipse. Let the outline in 
elevation intersect xy in a', a^ ; bisect a'a^ in c . Project 
a', a x ' and c on to the plan of the axis at #, a v and c. The 
horizontal trace of the cone consists of an ellipse of which C 
in the centre, and AA 1 the major axis. 

To set out the ellipse. First Method. Bisect the angle 
v'a'a^ by the line a'i' ; draw i'f perpendicular to xy, and 
with i' as centre and i'f as radius draw the circle shown ; 
this circle is the elevation of a focal sphere of the cone with 
respect to the ground as section plane (see Art. 76). There- 
fore the point of contact/' is the elevation of one focus of 
the ellipse, and fi, projected from/', is its plan. Set off a.f x 
equal to af v then f x is the second focus. The minor axis 
of the ellipse may now be found and the ellipse drawn, 
as explained in Chapter IV. 

Second Method. Through c, the middle point of a'a^, 
draw h'k' perpendicular to v'e . On h'k' describe a semi- 
circle, and draw e'b ' perpendicular to h'k'. Then c'b ' is the 
length of the semi-minor axis, and the horizontal trace of 
the cone can now be completed as before. 

To explain this construction, observe that c', being the 
mid-point of a x 'a, must be the elevation of the minor 
axis, the minor axis itself being a chord of that circular 
section of the cone which has h'k' for its elevation. One- 
half of this circle is turned about HK until it is parallel 
to the vertical plane ; its elevation then being the semi- 
circle on lik' ; the semi-minor axis appears as c'b '. 

Third Method. Repeat the construction of the second 
method for a number of points in a'a x ' in addition to the 



xii r 



CURVES AND CURVED SURFACES 



32: 








mid-point c . The lines corresponding to c'b^ thus obtained 
will be ordinates of the elliptic trace, and may be set off on 
each side of va v on the projectors from the respective 
points in a'a(. This method is useful when the cone is so 
situated that its trace is a parabola or hyperbola; or an 
ellipse so elongated that the end A x is inaccessible. 

Fourth Method. Draw the plan of the focal sphere or 
any other sphere inscribed in the cone, and from v draw 
tangents to this circle. These tangents give the plan of the 
cone in outline and must touch the trace (see Art. 284). 
The trace can therefore be drawn, since it is an ellipse, of 
which the major axis and a tangent are known (see Prob. 98). 



324 PRACTICAL SOLID GEOMETRY chap. 

275. Problem. To determine the horizontal trace 
of a given cylinder of indefinite length, whose axis ED 
is parallel to the vertical plane. 

The cylinder may be given by its diameter, the position 
of a point D in its axis, and the inclination 6 of the axis ; 
or by two inscribed spheres as explained in Art. 273. In 
either case the plan and elevation of the outline of the 
cylinder are readily determined. Let them be as shown in 
the figure. 

The horizontal trace of the cylinder, being a plane 
section, must be an ellipse. To determine the plan of the 
ellipse we may proceed as follows : 

Let the outline in elevation meet xy in a'a t ' ; bisect 
a'aJ in /. Project a, a' v c on to the plan of the axis at 
a, a v c, and let the projector from c intersect the plan of 
the outline in b, b y Then c is the centre, and aa v bb : the 
major and minor axes of the elliptic trace. The ellipse can 
be drawn by any of the methods described in Chapter IV. 

The elevation of the trace is aa^. 

The foci of the ellipse are shown in the figure ; these 
might have been determined by a focal sphere, as was 
explained for the cone in the last problem. 

Examples. 1. A circle |" diameter, centre s, is the plan of a sphere 
which rests on the ground. A point v, 1" from s, is the plan 
of the vertex of a cone of indefinite length which circumscribes 
the sphere, the height of /"being iV'. Draw the plan of the 
cone and an elevation on a vertical plane parallel to VS. Set 
out the horizontal trace of the cone. 

2. A cylinder, 2" diameter, has its axis parallel to the vertical 

plane and inclined 50 to the ground. Determine its horizontal 
trace. 

Show the elevation of the section made by a vertical plane 
parallel to, and distant i-" from the axis. 

3. Draw a triangle a! b'c with b'c' in xy ; make b'c' 3", a'b' 4", 

and a'c 2^". This is the elevation of a cone, the plan of whose 
axis makes 40 with xy. Determine the horizontal trace of 
the cone. See Prob. 276. 

4. A cylinder 2" diameter has its axis inclined at 45 to the ground 

and 30 to the vertical plane. Determine its horizontal trace ; 
also its vertical trace. See Prob. 277. 



XIII 



CURVES AND CURVED SURFACES 



525 




x a. 



b 




c 



\CL 



f ! 



a 



k 



5. Determine the trace of the cone of Ex. 2, p. 330, on a horizontal 

plane J" above the ground. 

6. Determine the section of the cone of Ex. 3, p. 330, by a hori- 

zontal plane which touches the smaller sphere at its highest 
point. 

7. A cone, vertical angle 6o, has its axis horizontal and 2" above 

the ground. Determine the section by a vertical plane (a) 
which makes 45 with the axis and is distant 2" from the 
vertex ; (/>) which makes 20 with the axis, and is distant A" 
from the vertex. 

8. Determine the vertical trace of a cylinder, ii" diameter, whose 

axis makes 40 with each plane of projection. 



326 PRACTICAL SOLID GEOMETRY chap. 

276. Problem. Having given mVn', the elevation of a 
cone, and also the projections of the axis EV, to determine 
the horizontal trace of the cone. 

Let it be observed that m'ri is not the true length of 
the major axis, nor is it the length of the elevation of the 
major axis of the required elliptical trace, for the axis of the 
cone is not parallel to the vertical plane as in Prob. 274. 
None of the methods there explained can be immediately used. 

First Method. Obtain an auxiliary elevation of the cone 
as seen when looking in the direction shown by the arrow 
in plan, that is on x'y' taken parallel to ev. In determining 
this elevation in outline use may be made of an inscribed 
sphere, as explained in Art. 273. Then any of the methods 
in Prob. 274 may be employed. 

Second Method. Bisect the angle vi n'v by the line n'i' 
and draw i'f perpendicular to xy ; then / is the centre of 
one focal sphere, and F is the corresponding focus. Bisect 
m'ri in c, and by projection from c and /' obtain c and_/ on 
ve, produced if necessary. The point C is the centre of the 
ellipse. 

Draw the projector n'r\ then it is evident that n'r is a 
tangent to the required ellipse. 

Draw fh perpendicular to n'r, and with c as centre and 
ch as radius describe the arc ha to meet ev in a ; then a is 
one extremity of the major axis of the required ellipse, and 
c? 1 is the other, where ca x is made equal to ca. The minor 
axis can now be found as in Prob. 79, and the ellipse con- 
structed. The plan of the cone is completed by drawing 
tangents from v to the ellipse. 

277. Problem. Having given the projections of cylinder 
of indefinite length, to determine the horizontal trace of the 
cylinder. (No figure.) 

As in the last problem, the axis of the solid is to be 
taken inclined to both planes of projection, and the outline 
in plan and elevation may be drawn as tangents to the pro- 
jections of two inscribed spheres. 



XIII 



CURVES AND CURVED SURFACES 



327 





The construction to find the trace is exactly like that for 
the cone ; but in the case of the cylinder may be somewhat 
simplified, because the minor axis of the ellipse is known, 
being equal to the diameter of the cylinder. 



328 PRACTICAL SOLID GEOMETRY chap. 

278. Problem. The plan of a cone of indefinite length 
is given, the plan of the axis being indexed ; determine 
the index of p, the given plan of a point on the upper 
portion of the cone. 

First Method. Let mvn be the plan of the cone, and ev, 
with the indices attached, the plan of the axis. 

Draw xy parallel to ev, and project /, v . Describe a 
circle to pass through p and touch vm, vn (Prob. 61); 
let its centre be s. First regard this circle as the plan of 
an inscribed sphere and determine its elevation, that is, the 
circle with s' as centre. Complete the elevation of the 
cone in outline by drawing the tangents v'k', v'k' . 

Now regard the circle, centre s, as the plan of a vertical 
cylinder, the elevation of which is indicated. Thus, the 
cone and cylinder circumscribe the same sphere, and hence 
(see Art. 273) their curve of intersection consists of two 
ellipses. But it is evident, from considerations of symmetry, 
that we shall obtain an edge view of these ellipses when 
looking horizontally in a direction at right angles to both 
axes ; and since /', ze/, z, u' are clearly the elevations of 
points which are on both surfaces, it follows that t'w and 
u'z are the elevations of these ellipses. 

Now P is on the surface of the cone, and it has been 
arranged that it is also on the cylinder, therefore it is on one 
or the other of the two ellipses ; but from the condition that 
P is on the upper half of the cone, it will be seen that P 
is on the ellipse TW, and hence p' is on t'w'. The required 
height of Pis therefore/'/, from which the index of P may 
be measured. 

See over leaf for the second method. 

Note 1. It should be observed that /Ms not on the sphere, centre S, 
although p is on the plan of this sphere. Draw k'k' joining the points 
of contact of the tangents from v to the circle, centre / ; join p'v' 
intersecting k'k' in /', then T is the point of contact of the generator 
PV with the sphere, centre S. Hence if p'c' be drawn parallel to t's', 
then C will be the centre of that inscribed sphere which contains P ; 
c is the plan of the centre of this sphere. This will be required in 
some future problems. 



XIII 



CURVES AND CURVED SURFACES 




330 PRACTICAL SOLID GEOMETRY chap. 

Second Method. Let mvn be the given plan of the 
cone, egJ 1A being the indexed plan of its axis. 

Draw any circle, centre s, to touch mv and nv ; this is 
the plan of an inscribed sphere. 

Draw vp, and regard it, not only as the plan of the 
generator through P, but also as the plan of the vertical 
plane containing this generator. Such a plane cuts the 
sphere in a circle, diameter ao, to which the generator PV 
is a tangent. 

An elevation of this circle and tangent are shown on xy, 
taken parallel to pv. In determining this elevation, the 
heights of e and v are known from the given indices : s' is 
the elevation of the centre of the sphere, and also of the 
centre of the circle diameter ad. 

Draw the tangent v'f. Then a projector from/ to meet 

v't' produced in p\ will determine r'p' the height of P, and 

the index of/ is thus known. 

Note 2. It will be useful for the student to observe that the 
generator VP touches the sphere, centre S, at T, and by obtaining t 
from /', joining si and drawing pc parallel to st, we obtain the plan of 
the centre of that inscribed sphere which contains /'. This sphere 
will be required in some future problems. 

Examples. 1. Describe a circle i-j" diameter, centre 5 ; take a 
point v I J" from s ; these are the plans of an inscribed sphere 
and vertex of a cone whose axis is inclined at 25 to the ground. 
Take a point/ on the given circle, distant 2j" from v, and regard 
this as the plan of a point on the surface of the cone. Determine 
the height of /"above the ground, if that of the vertex be 2-\". 
Determine the plan of that inscribed sphere of the cone 
which passes through P. 

2. A cone whose vertical angle is 90 , rests with a generator on the 

ground ; draw its plan. Determine a point on the surface of 
the cone which is ii" high and 2" distant from the vertex. 

3. Two circles 2" and l|" diameter, which have their centres 1^" 

apart, are the projections of two spheres, the heights of the 
centres being respectively ii" and 1" above the ground. 
Regard the points of intersection of the circles as the plans of 
two points on the upper and lower surface of the cone which 
circumscribes the spheres. Required the indices of the points ; 
unit = 0.1". Also determine the inscribed spheres on the sur- 
faces of which the points lie. 



XIII 



CURVES AND CURVED SURFACES 




332 PRACTICAL SOLID GEOMETRY chap. 

279. Problem. The plan of a cylinder of indefinite 
length is given, the plan of the axis being indexed ; deter- 
mine the index of p, the given plan of a point on the 
lower half of the cylinder. 

First Method. Let cd, with the indices attached, 
be the plan of the axis of the cylinder. Take xy parallel 
to cd, project cd', and draw the outline elevation of the 
cylinder. 

Describe a circle to pass through p and touch the out- 
line of the given plan of the cylinder. First regard this 
circle as the plan of a sphere inscribed in the given cylinder, 
and then regard it as the plan of a vertical cylinder, the 
elevation of which may be at once drawn. In this way it is 
ensured that the two cylinders circumscribe the same sphere, 
and therefore intersect each other in two ellipses. From 
what was said in the last problem, it will readily be seen 
that t'w' and u'z are the edge elevations of these ellipses, 
and that p' will be on u'z , since P is on the lower half of 
the cylinder. Measure p'r , and index / accordingly. 

Note i. Since any number of points may be taken on the plan of 
the vertical cylinder in either this problem or the last, and the corre- 
sponding elevations determined, it follows that the projections of any 
number of generators of the given cone or cylinder may be deter- 
mined ; and by finding their horizontal traces the horizontal trace of 
the cone or cylinder may be determined, as the fair curve through 
these points. This constitutes a method of setting out the trace of a 
cone or cylinder, additional to the methods given in Probs. 274 
to 277. 

Second Method. By regarding a cylinder as the 
particular case of a cone when the latter has its vertical 
angle diminished indefinitely, it will be seen that the second 
method of the previous problem (p. 330) will apply to the 
case of a cylinder, the generator through P being now 
parallel to the axis. 

Note 2. The sphere inscribed in the cylinder and containing P 
may be determined as in Note 2 of the last problem. 



XIII 



CURVES AND CURVED SURFACES 



333 





s 



\ 




Examples. 1. Draw a line ab i^" long ; a w b 22 is the indexed plan 
of the axis of a cylinder 2" diameter. Select any point/ j" 
from ab this is the, plan of a point on the surface of the 
cylinder. Required the index of the point p. Unit = 0.1". 
Also determine the indexed plan of the inscribed sphere which 
passes through P. 

2. Determine a sectional elevation of the cylinder of Ex. 1 on 

a vertical plane whose horizontal trace makes 60 with ab. 

3. Copy the above figure double size ; then draw the elevation of the 

ellipses TIV, UZ on a vertical plane which makes 45" with xy. 



334 PRACTICAL SOLID GEOMETRY chap. 

280. Problem. The plan of a cylinder of indefinite 
length is given, gd, the plan of the axis, being indexed. 
Determine the true shape of the section made by the 
vertical plane, the horizontal trace of which is lm. 

Bisect the angle nqr by the line qi x ; draw /,/ at right 
angles to lm, then I x is the centre of a focal sphere with 
regard to the section plane LM, and F x is the corresponding 
focus. From symmetry F is the other focus where rf= qf v 

Draw xy parallel to lm and obtain g ', d' by projection 
from g, d, setting off the heights given by the indices. 

Now since LM is a vertical plane, the radius L l F l of the 
sphere is horizontal ; and I x is on the axis of the cylinder. 
Hence z' a ' and /' coincide, and each is on g'd' . 

Project/' and/' from/ and/ Bisect//' at right angles 
by the line b'by, and make c'b' = c'b{ = radius of cylinder; 
then b'b^ is the minor axis. The major axis a'a^ may now 
be determined and the ellipse may be completed by any of 
the methods explained in Chapter IV. 

281. Projection of any surface of revolution. Reasoning 
as in Art. 273, it is readily seen that a series of spheres can be 
inscribed in any surface of revolution. The surface itself 
may be conceived as generated by, or the envelope of, its 
inscribed spheres. It is shown in pure geometry that the 
outline projection of a surface of revolution on any plane is 
the envelope of the projections of its inscribed spheres ; 
that is, the projection may be generated by a circle of vari- 
able diameter moving with its centre on the projection of 
the axis of the solid. 

In illustration, refer to the figure on page 337. The 
axis CD of the surface of revolution is parallel to the 
vertical plane, and the elevation consists of two circular 
arcs, having c'd' as a common chord. The plan of the 
surface might be determined thus : 

First draw a series of circles inscribed in the elevation ; 
then draw the plans of the spheres represented by these 
circles ; then draw the envelope, or the curve which 
touches all these plans. 



XIII 



CURVES AND CURVED SURFACES 



335 




The method of clr wing the plan which is adopted in 
the next problem is fuller and more instructive. The curve 
HPRG on the surface which projects into the outline in 
plan is determined, and a number of points in the plan are 
obtained. This gives more definiteness than would be the 
case if the method of envelopes alone were used. 



336 PRACTICAL SOLID GEOMETRY chap. 

282. Problem. A surface of revolution is generated 
by a given circular arc CTD, which revolves about its 
chord CD. It is required to determine the projections of 
the surface when the axis CD is parallel to the vertical 
plane and inclined at e to the horizontal plane. 

For the elevation draw c'd' making an angle 9 with xy, 
and complete the elevation by describing the two arcs on 
c'd'. Pet o be the centre of one of these arcs. 

To obtain the plan. Suppose a vertical projector to move tan- 
gentially round the surface ; then the foot of this perpendicular will 
trace the outline of the plan. The projector during its circuit will 
touch the surface along a curve, the plan of which will be identical 
with the plan of the outline. We shall determine this curve, employing 
inscribed spheres for the purpose. 

From the centre o draw any radius o'b' , intersecting c'd 
in / ; with centre s draw the circle through //, draw b'a per- 
pendicular to c'd' , and e's'f parallel to xy. The lines a'b' 
and e'f intersect in />'. 

Project from c'd' and thus obtain the plan cd, which is 
parallel to xy. Project s from s'. With centre s, radius 
s'f, describe a circle, and draw the projector from p' to 
cut this circle in p, p. Then p, p are points on the 
required plan, and p' is the elevation of the point where 
the projector touches the surface. 

For let S be the centre of an inscribed sphere. Then a'b' is the eleva- 
tion of its circle of contact. Let e'f represent a circle on the sphere. 
These circles intersect in P. Now the vertical projector through 
P touches the surface at P, because it evidently touches the sphere, and 
in the immediate neighbourhood of P the two surfaces coincide. 

Repeat the above construction for other inscribed spheres. 
In particular, draw the radii o't' and o'g' , the latter being 
parallel to xy, and obtain the important points R and G, 
and by symmetry H. 

If a curve be drawn as shown through h', p', r, g', and 
other similar points, it will be the elevation of the curve 
on the given surface of revolution, such that the outline plan 
of the surface is identical with the plan of the curve. 

Examples. 1. Work Prob. 282 when c'd' = 3 V', o'f = 2%", 6 = 50 . 
Add an auxiliary elevation of the surface on a vertical plane 
which makes 45 with xy. 



XIII 



CURVES AND CURVED SURFACES 



337 




A surface generated by the revolution of a circular arc about an 
external line is 3" long, i|" diameter at each end, and g" 
diameter in the middle. Draw the plan when its axis is in- 
clined at 45. Draw also the elevation on a vertical plane 
making 30 with the plan of the axis. 

An annulus is generated by a sphere \\ diameter, whose centre 
moves round a circle 2.\" diameter. Draw the plan of the 
surface when the annulus rests on a plane inclined at 45. 
Hint. The plan is similar in shape to Fig. 123. 



338 PRACTICAL SOLID GEOMETRY chap. 



283. Miscellaneous Examples. 

1. The plane of a circle of 2" diameter is inclined at 50 . A 
diameter of this circle is inclined at 30 . Draw the plan of 
the circle and also an elevation on a plane parallel to the 
inclined diameter. (1893) 

*2. Fig. (a). The side elevation of a circular tin canister with the lid 
(hinged at A) partly open is given. Draw its plan, and an 
elevation on a vertical plane, having xy as ground line. (1885) 

3. VTH\s an oblique plane, the traces vt and th making angles of 

6i and 71 with xy. A point c, 2.2" from / and 1.7 from xy, 
is the plan of the centre of the base of a right circular cone 
which has a generator in the horizontal plane and its base in the 
given plane VTH. Determine the plan of the cone. (1889) 

4. A sphere of radius 1.5" resting on the ground is cut by a plane 

inclined at 60", whose horizontal trace touches the plan of the 
sphere. Draw the plan of the section. (1877) 

5. A vertical cylinder 2" in diameter is cut by a plane inclined at 

50 , and having a horizontal trace which touches the plan of 
the cylinder and makes 35 with the ground line. Draw the 
elevation and development of the curve of section. (1S77) 

6. Fig. (b). A horizontal right cylindei lies on a truncated right cone 

as shown, the axis of the cylinder passing through that of the cone. 
Draw the figure full size, and make a sectional elevation of the 
two solids on a vertical plane distant A-" from the axis of the 
cone, and making 50 with the axis of the cylinder. (1887) 

*7. Fig. {e). v'a'b' is the elevation of a right cone. A circle, whose 
centre is 0, is drawn on the elevation, touching v'b' ; and two 
tangents to the circle, c'd', c'e', are drawn. Develop the cone, 
opening it along the line VB, and determine the developments 
of the circle and lines. (1895) 

*8. Fig. ((-). Find the real length of the shortest line that can be 
drawn on the right cone, whose plan is given, joining the points 
whose plans are/, q. Height of cone 3V. (1878) 

9. The axis of a cone is inclined at 6o to the H.P., the apex point- 
ing upwards and to the right ; the generatrix makes an angle 
of 25 with the axis. Determine a section of the cone, the plan 
of which will be a circle of 3" diameter. (!897) 

*10. Fig. (/). The elevation of a right cone ; and the plan of the 
axis are given ; draw the plan of the cone. (Honours, 1880) 

*11. Fig. (d). A surface of revolution is shown in elevation ; draw 
a. plan on x , y . 
12. A cone, base 2.70" diameter, height 2.35", has its axis inclined 
at 40. A curve is traced on the cone, which, in development, 
would be a circle of 1" radius touching the base of the cone. 
Draw the plan of the cone, and of the curve traced on it, 
touching the base of the cone at its highest point. (1894) 



XIII 



CURVES AND CURVED SURFACES 



339 




CHAPTER XIV 

TANGENT PLANES TO SURFACES 

284. Nature of the contact of a tangent plane. Let any 
three non-collinear points, A, B, C, Fig. (a), near to one 
another, be taken on the curved surface of a solid figure, 
and suppose a section of the figure to be made by the plane 
through A, B, C. If the surface in the neighbourhood be 
wholly convex (or wholly concave), like that of a sphere or 
the rounded portion of a vase, the section plane will inter- 
sect the surface in a closed curve of some kind passing 
through the three points ; this is illustrated in Fig. (a). 
Consider now any point P on the surface and within this 
curve, and suppose the points A, B, and C to approach 
indefinitely near to P. Then the plane ABC has a definite, 
limiting position, and is called the tangent plane to the sur- 
face at P. The tangent plane is said to touch the surface 
at P. 

Let P be projected on the section plane at /. These 
two points are too near together to be distinguished separ- 
ately in the figure, and ultimately coincide. Now any line 
in the plane ABC through/ will intersect the curve ABC, 
and therefore the surface, in two points, shown at D and E 
in the figure. Again suppose A, B, C to approach indefinitely 
near to P, so that in the limit DE becomes a line through 
P in the tangent plane ; hence this line meets the surface in 
two consecutive points, that is, N touches the surface or is 
tangential to it. We thus see that any line in a tangent 



chap, xiv TANGENT PLANES TO SURFACES 



341 




(*) 




plane which passes through the point of contact touches 
the surface at the point. Thus we have a series of tangent 
lines at any point P on a surface, which all lie in the 
tangent plane at P. The tangent plane at P may in fact 
be looked on as having been generated by a tangent line 
which is rotated about P so as to always touch the surface at 
P. The axis of rotation is the normal to the surface at P. 

Again, suppose any plane through p, inclined to the 
plane ABC, to cut the latter in the line DP, and the surface 
in the curve DPP. In the limit this line and curve will 
both pass through P, and the line will touch the curve at P. 
It is so important that this theorem be well understood 
that we present it in several different forms. 

Thus, if any plane section of a surface be taken through a 
point P on the surface, then the line in which the tangent plane 
at P is cut by the section plane is a tangent at P to the curve 
of the section. 

Or, if a plane touch a surface at P, then the traces of the 
plane and the surface 011 any second plane through P also 
touch one another at P. 

This theorem may be generalised and stated in the 
form : If two surfaces touch one another at P, then the traces 
of the surfaces on any other surface through P will also touch 
o?ie another at P. 

As a simple example, suppose a cone to stand with its 
circular base on the ground, then the horizontal trace of 
any tangent plane will touch the base. 

The properties of tangent planes and lines above stated 



342 



PRACTICAL SOLID GEOMETRY 



CHAP. 



have been illustrated for the case of a surface wholly convex 
or wholly concave in the neighbourhood of the point of 
contact P. They are equally true for a surface which at P 
is co?ivex in some directions and concave in others, like a horse's 
saddle, or the hollow neck of a vase ; but in this case the 
surface around /'lies partly on one side of the tangent plane 
and partly on the other, and the tangent plane at P cuts 
the surface in a curve which has a node at P, Fig. (b), the 
two branches having points of inflexion where they cross. 
The surface near P is divided into four regions by the two 
branches of the curve of intersection. Two opposite regions 
lie on one side of the tangent plane, and the other two on 
the other side. 






285. Surface of revolution with inscribed cone and 
sphere. Let QR be any plane curve, and VC any line in 
its plane. Let PV and PC be the tangent and normal at 
P. With centre C, describe the circle through P. 

Now let the figure rotate about 
VC as axis. Then the cone and 
sphere described by PV and the 
circle are tangential to the surface 
p? generated by QR, the contact ex- 
7\ tending along the circle traced by 

P, and represented in the figure by 
PP' perpendicular to the axis. The 
cone and sphere are said to be in- 
scribed in (or they may circum- 
scribe) the surface. The tangent 
plane to one of the surfaces, at any point in the circle of 
contact PP\ also touches the other two. 

The surface of a cone or cylinder is generated by the 
motion of a straight line, and the tangent plane at any 
point of either touches the surface along the line or 
generator through the point, and hence touches all in- 
scribed spheres. 

See Theorems 23 to 48, Appendix II. 




XIV 



TANGENT PLANES TO SURFACES 



343 



286. Problem. To determine the traces of the plane 
which shall be tangential to a given cone, axis vertical, 
vertex V, and shall pass through a given point P on its 
surface. 

The projections of the cone are shown in the figure. 

Draw vb, z/b\ the projections of the generator through 
P, and at b draw 

the tangent nm \ 

perpendicular to 
vb ; this is the 
horizontal trace 
of the required 
tangent plane. 

Through p 
draw pc parallel 
to nm, and p'c' 
parallel to xy. 
These are the 
projections of a 
horizontal line 
passing through 
P, lying in the 
tangent plane, 
and terminated at 
C by the vertical 
plane. 

Hence Im 
drawn through 
c and m is the 
required vertical trace. 

Examples. 1. Draw the plan and elevation of a cone with its 
base on the ground, diameter of base if", height 2^". Draw 
the traces of a plane which touches the cone along a generator 
whose plan makes 45 with xy. 

2. Draw the plan and elevation of a line inclined at 45 to touch the 

cone in Ex. 1 at the middle point of the given generator. 

3. Determine the traces of a tangent plane as in Ex. 1, but with 

the cone inverted so that its vertex rests on the ground. 




344 PRACTICAL SOLID GEOMETRY chap. 



287. Problem. A given cone, vertex V, rests with its 
base on the horizontal plane, determine the traces of a 
plane which shall pass through a given external point P 
and be tangential to the cone. 

The required plane must contain the line VP. Therefore 
obtain h, the plan of the horizontal trace of VP. 

Draw nhm tangential to the plan of the base of the cone. 
Then nm is the horizontal trace of the required plane. 

Determine r the elevation of the vertical trace of VP. 
Then r'lm is the required vertical trace. 

Note. Since two tangents may be drawn from h to the plan of the 
base of the cone, there are two tangent planes satisfying the given 
conditions. 

288. Problem. To determine the traces of the plane 
which shall be tangential to a given sphere, centre S, at 
a point on its upper surface, the plan p of the point being 
given. 

First Method. With centre s describe a circle to pass 
through p, and obtain a'b' the elevation of this circle ; a 
projector from /> to meet a'b' will determine/'. At d and 
b' draw tangents to the circle, meeting each other in v and 
xy in e',f; th&nv'e'f' is the elevation of a cone which 
touches the given sphere along the horizontal circle through 
P. The plan of the cone is easily obtained. 

Now determine by Prob. 286 the plane LMN to touch 
the cone at P ; this will be the required tangent plane. 

Second Method. The tangent plane at Pis perpendicular 
to SP, hence its traces are perpendicular to sp and s'p' re- 
spectively. 

To determine these traces draw pc perpendicular to sp 
and/V parallel to xy ; then PC is a horizontal line passing 
through P, contained by the tangent plane and having C 
in its vertical trace. Hence Im drawn through c perpen- 
dicular to s'p' will be the vertical trace, and mn perpen- 
dicular to sp will be the horizontal trace of the required 
tangent plane. 



XIV 



TANGENT PLANES TO SURFACES 



345 



X 




346 PRACTICAL SOLID GEOMETRY chap. 



289. Problem. A given cylinder, axis AB, lies with 
a generator on the horizontal plane. It is required to 
determine the plane which shall touch the cylinder at a 
point whose plan p is given. 

Let the rectangle cdefbe the given plan of the cylinder. 

First Method. Conceive the tangent plane in position ; 
it will touch the cylinder along the generator through P. 
To a person looking horizontally in the direction of the 
arrow in plan, the cylinder will appear as a circle, and the 
tangent plane as a line touching this circle. This view 
will now be drawn. 

Take x'y perpendicular to ab, and draw the elevation 
of the cylinder, that is, the circle with centre a". Obtain 
p" by projection from /, and at/" draw the tangent v"o per- 
pendicular to d'p" . This is the edge elevation of the tangent 
plane, and on drawn through o parallel to ba is the horizontal 
trace. The vertical trace hn may be found as in previous 
problems, since the horizontal trace and the projections of 
a point P in the plane are known. 

Second Method. Draw the projections of a sphere in- 
scribed in the cylinder and passing through P; then deter- 
mine the tangent plane to the sphere at P (Prob. 288). 
This will be the required tangent plane (see Art. 285). 

Note. If on be nearly parallel to xy, the method of Prob. 211 
may be used to determine the vertical trace, as here indicated. 

Examples. 1. Draw the plan and elevation of a cone with its 
base resting on the ground, diameter of base i|", height 2^", 
axis 2" from the vertical plane. A point P is 2" to the right of 
the axis of the cone, Ij" above the ground, and i|" from the 
vertical plane. Draw the traces of the two planes which pass 
through P and touch the cone. 

2. Draw the projections of the line from P to touch the cone, so 

that the length from P to the point of contact is the least 
possible. 

3. A sphere i\" diameter has its centre f" above the ground, and 

2\" in front of the vertical plane. A point P on the upper 
half of the surface is i|" above the ground, and 2" in front of 
the vertical plane. Draw the traces of the tangent plane to 
the sphere at the point P. 



XIV 



TANGENT PLANES TO SURFACES 



347 




Draw the projections of the tangent line to the sphere at the 
point P in Ex. 3, such that it is inclined 30 to the ground. 

A cylinder iV' diameter touches the ground along a generator 
which makes 30 with the vertical plane ; determine the traces 
of a plane which touches the cylinder at a point P, ij" above 
the ground. 

Draw a line through the plan of P in Ex. 6, making an angle 
of 6o with xy and 30 with the plan of the axis. If this is the 
plan of a line which touches the cylinder at P, draw the eleva- 
tion. 



348 PRACTICAL SOLID GEOMETRY chap. 

290. Problem. To determine the traces of a plane 
which shall touch a given cone, vertex V, resting with its 
base on the ground, and a given sphere, centre S. 

Circumscribe the given sphere by a cone, vertex A, so that 
the two cones are similar and similarly placed. The pro- 
jections of this circumscribing cone are shown in the figure. 

Draw mn an externa/ common tangent to the plans of 
the bases of the two cones. This is the horizontal trace of 
a required tangent plane. 

Determine the projections of C, the vertical trace of the 
line VA. Then hn drawn through ;;/ and c is the required 
vertical trace. 

For, any plane which touches the two cones must touch 
the given cone and sphere ; and since the plane contains the 
two vertices its vertical trace passes through the vertical 
trace of VA. 

Note. It will be observed that the plane LAIN is such that the given 
sphere and cone lie on the same side of it. There is another such 
plane the horizontal trace of which is the other external tangent to the 
circles v and a as shown. 

There is a second pair of planes each of which passes be- 
tween the given sphere and cone. The manner in which the 
traces are obtained is exhibited in the same figure, in which 
an inverted cone is taken to circumscribe the given sphere, 
the base angles of this cone and the given cone being 
equal. The trace of the inverted cone on the horizontal 
plane is the small circle shown in plan. The internal 
common tangents to this circle and the one with centre v 
will be the horizontal traces of the two tangent planes, the 
vertical traces beinc: found as in Prob. 211. 



*& 



Examples. 1. Draw the plan and elevation of a cone with its 
base on the ground, axis 3" from the vertical plane, diameter 
of base 2", height 3". A sphere ij" diameter has its centre 
2.y to the right of the axis of the cone, ij" from the vertical 
plane, and 1" above the ground. Determine the traces of one 
tangent plane to the sphere and cone, and draw the horizontal 
traces of all the tangent planes. 



XIV 



TANGENT PLANES TO SURFACES 



349 



V 



a 4 



x 




c 



77jy 



>//l 



V 



'a 



T s; 



rt 



2. Draw the projections of any line which touches the cone and 

sphere in Ex. i. 

3. Draw the traces of a plane which shall touch the cone in Ex. I, 

page 346, and be J-" from the given point P. 

4. Two points A and B are respectively 2" and 1" from each of 

the planes of projection. AB is 2f" long. Determine the 
traces of a plane through A inclined at 55 to the ground, and 
distant f" from B. 

5. Determine a sphere which shall contain the point A, Ex. 4, be 

distant |" from B, and shall make an angle of 55 with the 
vertical plane. 

6. Determine a plane which shall touch a sphere, 2" diameter 

whose centre is in xv, and shall make 30 with xy. 



350 PRACTICAL SOLID GEOMETRY chap. 

291. Problem. To determine the traces of a plane 
which shall be tangential to two given spheres, centres S 
and C, and shall have a given inclination o. 

Let the two spheres be each circumscribed by a vertical 
cone with base angle equal to the given inclination 6. 

These cones may be either upright or inverted, and by 
taking the four possible combinations eight common tangent 
planes may in general be obtained. 

In the figures the horizontal traces only are shown ; 
the vertical traces may be found as in the preceding 
problems. 

In (a) two upright cones are taken. 

In (i>) one cone is upright and the other inverted. 

In (c) both cones are inverted. 

In (d) one cone is inverted and the other upright. 

In determining the horizontal traces of the tangent 
planes care must be taken to draw the proper common 
tangents. Thus in (a) and (c) each tangent plane is such 
that the spheres are situated on the same side of it, hence 
the external common tangents must be drawn. 

On the other hand, in (b) and (d) each tangent plane 
passes between the spheres, consequently the internal 
common tangents must be drawn to give the traces. 

Note. Some of the tangent planes may be coincident or impossible, 
and the number of solutions may be anything from eight to none, 
according to the data. 

Examples. 1. The centres of two spheres which rest on the ground 
are 2^" apart ; the centre of one is 2" from the vertical plane, 
and that of the other is 1^". The diameters of the spheres 
are iiy"and 1" respectively. Determine the traces of a plane 
touching the spheres and inclined at 55. Draw the horizontal 
traces of all such tangent planes. 

2. Two points A, B are respectively 2" and ih" from each plane of 

projection and their projectors are 2^" apart. Determine a 
plane which shall be inclined at 50, and 1" distant from both 
A and B, and which shall lie below A and above B. 

3. Determine all the planes which are 1" distant from A and B, 

Ex. 2, and which make 50 with the vertical plane. 



XTV 



TANGENT PLANES TO SURFACES 



35i 



X 





y jl^ 





x- 





c\i 



-y 







y 



352 PRACTICAL SOLID GEOMETRY chap. 

292. Problem. To determine the traces of a plane 
which shall touch a given cone, axis VA inclined to both 
planes of projection, and have a given inclination e. 

Let r'v's' and uvw be the projections of the given cone. 
Draw the projections of the vertical cone which has its vertex 
at f^and its base angle equal to 0. 

Draw the projections of any sphere, centre C, inscribed 
in the given cone, and also the projections of a cone, vertex 
V v similar and similarly placed to the first cone and cir- 
cumscribing the sphere. 

Then n/n, an external common tangent to the plans of 
the bases of the two upright cones, is the horizontal trace of 
a tangent plane to these cones, and ml is its vertical trace. 
This plane passes through V, touches the sphere C, and so 
touches the ajven cone. And it is inclined at 6. 

Note I. The inclination 6 cannot be less than that of the least 
inclined generator of the given cone. 

Note 2. There are, in general, four planes satisfying the given 
conditions, two being obtained as shown, and two others from an in- 
verted cone (not shown) which circumscribes the sphere C. 

Note 3. Instead of the vertical cone through V we might have 
taken one circumscribing any second sphere inscribed in the given 
cone. 

293. Problem. To determine the traces of a plane 
which shall touch a given cylinder, axis AB, and have a 
given inclination e. 

Draw the projections of any two spheres, centres and 
C, inscribed in the given cylinder. 

Now draw the projections of the two upright cones 
which circumscribe the spheres, each cone having a base 
angle 9. Determine the plane NML to touch the cones. 

Since this plane touches the two cones, it touches both 
spheres without passing between, hence it must touch the 
cylinder ; and it is inclined at 6. 

N te 1. The value of 6 cannot be less than the inclination of the 
axis of the cylinder. 



XIV 



TANGENT PLANKS TO SURFACES 



353 



X r 




2 A 



354 PRACTICAL SOLID GEOMETRY chap. 

294. Problem. Having given the projections of a 
cone, axis VA, and p the plan of a point on its surface, to 
determine the traces of a plane which shall pass through 
P and touch the cone. 

Let r'v's and uvw represent the given cone. 

First Method. By Prob. 278 determine/' and also the 
projections of T (not shown), the point of contact of PV 
and any sphere inscribed in the cone. 

The tangent plane to the sphere at the point T is the 
required plane ; this may be determined by either of the 
methods in Prob. 288. 

Second Method. Determine p' as in the preceding 
method, and find Q, the horizontal trace of -PV. 

If the horizontal trace of the cone be determined, that 
is to say, an ellipse whose elevation is r's', the tangent 
drawn at q to this ellipse will be the horizontal trace of the 
required tangent plane (Art. 284). 

But it is unnecessary to actually draw the ellipse. Deter- 
mine the foci,/ and f x in plan, as in Prob. 276. 

Join qf and qf v and bisect the angle exterior to fqf v 
Then the bisector oh touches the elliptic trace at // and is 
the horizontal trace of the required tangent plane. 

The vertical trace may be determined as in the preceding 
problems. 

Examples on Problems 292 to 294. 

1. The plan and elevation of the axis of a cone make angles of 30 

and 45 with xy ; the vertex is 2-J-" above the ground and 
ij" in front of the vertical plane. The sphere which is in- 
scribed in the cone and rests on the ground is ij" diameter. 
Determine the traces of a plane which touches the given 
cone and is inclined at 6o to the ground. 

2. A sphere 1^-" diameter has its centre l" above the ground and 

2-J" in front of the vertical plane. A point A, 2" to the right 
of the centre of the sphere, is 1" in front of the vertical plane 
and 3" above the ground. Determine the traces of a plane 
through A which shall touch the sphere and make an angle of 
65 with the ground. 

3. The plan and elevation of the axis of a cylinder 2" diameter 



XIV 



TANGENT PLANES TO SURFACES 



355 




4. 



make angles of 25 and 40 with xy. Determine the traces 
of a plane which shall touch the cylinder and have an inclina- 
tion of 65. Draw the horizontal traces of all the planes which 
satisfy the given conditions. What are the least and greatest 
possible inclinations of the planes which touch this cylinder? 

Take the cone Ex. I. Select a point p i|" to the left of the plan 
of the vertex, and 2" from xy. This is the plan of a point 
on the upper half of the cone. Determine the traces of the plane 
which touches the cone at P. 

Determine a plane which shall touch the cylinder of Ex. 3 along 
a line whose plan is A-" distant from the plan of the axis. 

Determine all the planes which touch the cylinder of Ex. 3 and 
make 6o with the vertical plane. 



356 PRACTICAL SOLID GEOMETRY chap. 

295. Problem. To determine a plane which shall 
contain a given line AB and touch a given sphere C. 

First Method.- It is obvious that the required plane will 
touch any cone circumscribing the given sphere and having 
its vertex in AB. Let the particular cone with vertex Vbe 
taken, where v is determined by drawing cv parallel to xy. 
Such a cone has its axis VC parallel to the vertical plane. 

Project v from v, and draw the tangents v'r, v's' ; then 
r'v's' is the elevation of the cone. 

Draw the elevation of the vertical cyclinder which 
circumscribes the given sphere. 

The cylinder and cone intersect in two ellipses whose 
edge elevations are the lines e'd' and/'^"' respectively. The 
plan of each ellipse is the circle centre c, since both ellipses 
are on the cylinder (see Art. 273). 

Produce e'd' to meet a'b' in ?i and xy in /; then TLM is 
an inclined plane which cuts the cone in an ellipse, the eleva- 
tion of which is e'd'. 

Conceive the tangent plane in position ; it intersects the 
plane TLM in a line which passes through iVand touches 
the ellipse whose elevation is e'd' (Art. 284). The plan of 
this line is nz (or nk), touching the circle which is the pro- 
jection of the ellipse. 

Therefore the required tangent plane contains the lines 
NZ (or NK) and AB, the horizontal traces of which are 
h and 0, in plan. Hence ho is the horizontal trace of one 
tangent plane satisfying the conditions of the problem. 
The vertical trace is found as in previous problems. 

Note 1. The point of contact of tangent plane and sphere is the 
point P where VZ meets the circle of contact of cone and sphere. 

Note 2. If the tangent NK be taken instead of NZ another 
tangent plane is found. Thus there are two solutions. 

Note 3. The inclined plane through fg' might have been taken 
instead of the one through e'd! , but this would not have met AB 
within convenient limits. 

Note 4. If the position of AB be such that it is inconvenient to 
determine V, a new elevation of the line and sphere may be drawn and 
the same method applied. 



XIV 



TANGENT PLANKS TO SURFACES 



357 



tf\ - r ' 




358 PRACTICAL SOLID GEOMETRY chap. 

Second Method. A modification of the first method is 
illustrated in the figure on page 359. The position of AB 
is altered but the same letters are used. 

Draw the elevation of the cone, vertex V, as before. 
Draw the traces ///;/ of the inclined plane which contains 
the circle of contact of cone and sphere. The given line 
intersects this plane in the point N. 

Suppose the tangent plane in position. It will intersect 
the plane TLM in a line which is a tangent from N to the 
circle whose elevation is e'd'. 

An auxiliary plan of the circle and the tzvo tangents 
from ./Vis drawn on x 1 y 1 taken parallel to //. Here ij = qc. 

The required plane contains NK or (JVZ) and AB ; its 
horizontal trace therefore passes through and // as before. 

Note 1. Since A' is in the tangent plane, the latter may be deter- 
mined by finding the traces of any two convenient lines through A' 
which meet AB. 

Note 2. The tangent plane touches the cone along the generator 
KV, hence it touches the sphere at a point P in KV. When P has 
been found the problem is reduced to Prob. 288. 

Third Method. Take a cone with its vertex at any 
convenient point in the given line and circumscribing the 
given sphere. 

Find the horizontal trace of this cone, and draw the two 
tangents to the trace from the horizontal trace of the given 
line. These two tangents are the horizontal traces of the 
required tangent planes. 

This method is the more straightforward to apply, but 
requires the setting out of the curved trace, which is a conic 
section. The other methods require only circles and straight 
lines to be drawn. 

Example. Draw a line inclined at 6o to xy, in which take a 
point a 1" from xy, and another point b 2^" from xy ; this 
is the plan of a line AB, whose ends A and B are 3" and 1" 
above the ground. Describe a circle Ij" diameter, with its 
centre 2" from both xy and ab ; this is the plan of a sphere 
whose centre is 2" above the ground. Determine the traces of a 
plane containing the line AB and touching the sphere. 



XIV 



TANGENT PLANES TO SURFACES 



359 




360 PRACTICAL SOLID GEOMETRY chap. 

296. Problem. To determine the traces of a plane 
which shall touch a given cone, axis VA, and pass through 
a given external point P. 

First Method. This problem may be at once reduced to 
the preceding one, since the required plane must contain 
the line joining the vertex of the cone to the given point, 
and touch any sphere inscribed in the cone. 

There will be two planes fulfilling the required condi- 
tions. 

Second Method. Let t'r's and vuw be the projections 
of the given cone, and p\ p those of the point. 

The required plane contains PV, hence its horizontal 
trace passes through the horizontal trace of PV ; therefore 
determine h, the plan of the horizontal trace of PV. 

The horizontal trace of the tangent plane must also 
touch the horizontal trace of the given cone, that is, an 
ellipse whose elevation is //. 

Determine the foci/j and /in plan, and the major axis 
of this ellipse, as in Prob. 276. 

On the major axis as diameter describe a semicircle ; 
and on hf x as diameter describe another semicircle, the two 
intersecting in n. 

Then /// drawn through n is a tangent to the ellipse, 
and is therefore the horizontal trace of a tangent plane. 
And It is the vertical trace, It being drawn through //, the 
elevation of the vertical trace of PV. 

Note. Two tangents can be drawn from h to the horizontal trace 
of the cone, hence there are two tangent planes which fulfil the 
required conditions. 

Examples. 1. Determine the traces of a plane which shall touch 
the cone of Ex. 1, page 354, and pass through a point P, 1" to 
the right of the vertex, 2" in front of the vertical plane, and 1" 
above the ground. 

2. The axis VA of a cone, vertical angle 25 , coincides with xy. A 
point P is 1 \" distant from both planes of projection, and the 
projector//' contains v, the projection of the vertex. Deter- 
mine a plane through P to touch the cone. Find also a plane 
which touches the cone and is inclined at 45 . 



XIV 



TANGENT PLANES TO SURFACES 



361 




362 PRACTICAL SOLID GEOMETRY chap. 

297. Problem. The projections of a cylinder, axis 
AB, being given, and p the plan of a point P on the 
surface of the cylinder, it is required to determine the 
traces of a plane which shall pass through P and touch 
the cylinder. 

First Method. The required plane will touch the 
cylinder along the generator through P. 

Determine/' by the method of Prob. 279, and draw the 
projections of the generator PD. 

This generator touches any inscribed sphere, centre C, 
at T, the projections of which may be determined at the 
same time that/' is found. 

The plane which touches the sphere at T is the required 
tangent plane, and may be found as in Prob. 288. 

Second Method. After having determined the projections 
of the generator PD, the horizontal trace nm of the tangent 
plane may be determined by first obtaining Q, the hori- 
zontal trace of PP>, and then drawing a tangent at q 
to the ellipse whose elevation is r s . 

This is done in precisely the same manner as was ex- 
plained in the second method of Prob. 294, Imn being 
the traces of the required plane. 

298. Problem. To find the traces of a plane which 
shall touch two given spheres and pass through a given 
point. (No figure.) 

This may be reduced to Prob. 296 by circumscribing 
both spheres by the same cone. The plane which passes 
through the given point and touches this cone must satisfy 
the given conditions. 

There are two such cones, one of which has its vertex 
between the spheres. Thus there are in general four 
tangent planes, two being found from each cone. 

299. Problem. To determine the traces of a plane 
which shall be tangential to three given spheres. 

Take the spheres in pairs, and determine the vertices 



XIV 



T ANCIENT PLANES TO SURFACES 



363 




V, IV of two cones which envelop any two of the three 
pairs ; then determine a plane which contains VW and 
touches any of the spheres. 

The projections of V and W are found by drawing the 
common tangents to the projections of the spheres, and 
the solution of the problem is then the same as that of 
Prob. 295. 



364 PRACTICAL SOLID GEOMETRY chap. 

390. Problem. To determine the traces of a plane 
which touches a given surface of revolution, axis vertical, 
at a given point P on its surface. 

First Method. The surface considered is the same as 
that in Prob. 282, d being the centre of the arc c'b'd'. 
One projection of /'being given, the other can be found by 
means of a horizontal section through the point, viz. dp b' 
in elevation and the dotted circle through/ in plan. 

Join db\ meeting c'd' in s . With / as centre and s'b' as 
radius describe a circle. This is the elevation of the in- 
scribed sphere which touches the given surface in the circle 
passing through P. 

And LMN, the tangent plane at P to this sphere, found 
as in Prob. 288, will also touch the given surface at P 
(Art. 285). 

Second Method. Draw the projections of a vertical cone 
touching the surface in a circle, of which db' drawn through 
p' is the elevation. Then P lies on the cone, and the plane 
touching the cone at P also touches the given surface. 

This tangent plane may be found as in Prob. 286. 

Examples on Problems 297 to 390. 

1. The plan and elevation of the axis of a cylinder 2" diameter 

make angle of 30 and 40 with xy. Determine the traces 
of a plane which touches the cylinder at a point whose plan 
is |" from the plan of the axis. 

2. Determine the traces of a plane which shall touch the two spheres 

of Ex. 1, p. 350, and pass through a point i^" vertically over 
the middle point of the line joining the centres of the spheres. 

3. Three spheres of 1.5", 1.0", and .4" diameters rest on the 

ground in mutual contact, and a fourth sphere .8" diameter 
rests on the three. Draw the plan of the group and determine 
the tetrahedron which envelops them, each face touching three 
spheres. 

4. Draw an equilateral triangle abc of 2" side, and with a, />, c as 
centres describe circles of 1", ", and ^" radii respectively. 
Take these circles as the plans of spheres which rest on the 
ground, and determine a plane to touch all the spheres. 

Hint. The horizontal trace of the tangent plane is found 
at once since the vertices of all the enveloping cones are on 



XIV 



TANGENT I'LANES TO SURFACES 



365 




the ground. The inclination of the plane is then readily deter- 
mined by taking an elevation on a vertical plane perpendicular 
to the horizontal trace. 

5. Take the surface of revolution of Ex. 2, p. 337, with its axis 

vertical and one end on the ground, and determine a plane 
which shall touch the surface at a point 2|" above the ground, 
the line joining the plan of the point to the plan of the axis 
making 45 with xy. 

6. A surface of revolution is shown in elevation at (f) on p. 485. 

Copy the figure double size and draw the plan. Determine a 
plane which shall touch this surface and make angles of 6o and 
40 with the horizontal and vertical planes of projection. 

7. In Ex. 6 determine the plan and elevation of the section of the 

capstan by the tangent plane, and show that the section has a 
node at the point of contact. 



*66 PRACTICAL SOLID GEOMETRY chap. 



301. Miscellaneous Examples. 

*1. Eig. (a). The plan of an inverted right cone with its vertex 
on the horizontal plane is given. The height of the plane of 
the base is 25. Determine the scale of slope of a plane con- 
taining the given point .Panel tangential to the cone. Unit o. 1". 

(1887) 

*2. Fig. (/;). The plans of two spheres are given and also the plan 
of a point P. Determine a plane touching both spheres, and 
passing through P. ( x 893) 

*3. Fig. (c). The given sphere rests on the horizontal plane. Deter- 
mine the scales of slope of planes touching it and containing the 
given line AB. Unit o. 1". (1884) 

*4. Fig. (d). ad is the plan and a the trace of a line inclined at 50 
to the horizontal plane. AB is the axis of a right cylinder of 
1.5" diameter and 1. 5" in length, its lower base resting on the 
horizontal plane. Draw the plan of the cylinder, and the hori- 
zontal trace of a plane tangent to it and inclined at 70 to the 
horizontal plane. ( 1S95) 

*5. Fig. (e). 9 ^ 8 is the axis of a right cylinder of 1 5 units diameter, 
and p.22 is the centre of a sphere of the same diameter. Draw 
a plane passing over the cylinder and under the sphere, and 
touching the surfaces of both. Show the point of contact be- 
tween the plane and the sphere. Unit = o. 1". ( X S96) 

*6. Fig. (_/). Determine a plane inclined at 65, making 70 with 
the given plane, and |" distant from the given point. Unit = 
0.1". (Honours, 1889) 

*7. Fig. (g). Two spheres are given, the index of the centre of the 
smaller being 11, and the lowest point of the larger being level 
with the highest point of the smaller. Draw a plane inclined at 
70 to the horizontal plane, and touching both the spheres. 
Unit = 0.1". (Honours, 1891) 

8. A sphere 1.75" in diameter touches both planes of projection. 

Determine the traces of a plane touching the sphere, and in- 
clined at 6o and 50 to the horizontal and vertical planes 
respectively. (1878) 

9. Three planes are mutually perpendicular, and each touches a 

sphere 1" diameter, which rests on the ground ; two of the 
planes are inclined respectively at 35 and 70 . Draw the 
horizontal traces of the three planes, the plan of their intersec- 
tions, and find the inclination of the third plane. 
10. A right cone, 2^" high, rests with its base (of ij" radius) on 
the H.P. A sphere of #'' radius touches it externally. Draw 
the true shape of the section of the cone, made by a plane pass- 
ing through the vertex of the cone, inclined at 75 to the H.P. 
and passing under, and touching the sphere. (1S98) 



XIV 



TANGENT l'LANES TO SURFACES 



367 




CHAPTER XV 



SURFACES IN CONTACT 



302. General remarks. Two surfaces which touch will 
have a common tangent plane and a common normal at the 
point of contact. The normal at any point on a sphere is 
the line joining the point to the centre ; and at any point 
P on the surface of a circular cone or cylinder, or any 
solid of revolution, is the line joining P to the centre C of 
the inscribed sphere which contains P. See notes to Prob. 
278. Any surface which touches the sphere at P also 
touches the surface of revolution at the same point. Thus 
problems on the contact of cones and cylinders may often 
be reduced to those on the contact of spheres, and thereby 
simplified. See Theorems 23 to 48, Appendix II. 

303. Problem. To determine the projections of three 
spheres, centres A, B, C, of given radii, which shall rest 
on the ground in mutual contact. 

Assume that the spheres with centres A and B are 
placed so that AB is parallel to the vertical plane. 

Describe circles, centres a and b', to touch xy and 
each other, their radii being those of the spheres A and B. 
These circles are the elevations of two of the spheres. 
Draw their plans with ab parallel to xy. 

Describe a circle (centre c ') to touch xy and the 
circle with centre //, the radius being that of the third 
sphere. Describe an equal circle to touch the one with a 



CHAP. XV 



SURFACES IN CONTACT 



369 




M^7 




as centre. Then inn and or are the lengths of the plans 
of BC and AC. Hence c, the plan of the centre of the 
third sphere, is a point of intersection of two arcs drawn 
one with centre a, radius or, and the other with centre b, 
radius mn. 

Draw c-[c parallel to xy to meet a projector from c in c . 
This gives the elevation of C, and the projections of the 
third sphere may now be drawn. 

The point of contact for each pair of spheres is indi- 
cated ; thus e is the intersection of b'c with a line through 
e x ' parallel to xy ; and e projected from c is in be. 

Example. Draw the projections of three spheres which touch 
each other and rest on the ground, the diameters being 2j", 
ij", 1". Show the projections of the points of contact. 

2 B 



37Q PRACTICAL SOLID GEOMETRY chap. 

304. Problem. To determine the projections of a 
sphere of given radius which shall touch (externally) a 
given sphere, centre S, at a given point P. 

The centre of the required sphere is on SJ 3 produced, 
therefore produce sp and s'p' . 

-Through p' draw the horizontal line a'b ' . 

Join s'b' and produce to c^, making c/b' equal to the 
given radius. With centre C-[ describe the circle through //. 

Draw CyC parallel to xy to meet s'p' produced in c, 
and project c on sp produced. With centres c and c, 
radius c-[b\ describe circles ; these are the projections of 
the required sphere. 

This construction is suggested by conceiving the re- 
quired sphere to roll on the given one until SC is parallel 
to the vertical plane, the point of contact remaining on 
the horizontal circle through P; their elevations while in 
this position can readily be drawn. The line c'c^ is the 
elevation of the path of C during this motion, and p'b' that 
of the point of contact. 

305. Problem.; To determine a sphere which shall 
rest on the ground and touch at a given point P a given 
sphere, centre S, which also rests on the ground. 

Through /' draw p'b' parallel to xy. Join s'b', and on 
s'b' produced, determine /, the centre of a circle which 
touches xy and the circle centre / (see Prob. 73) 

Draw c/c parallel to xy to meet s'p' produced in c ; then 
c is the elevation of the centre of the required sphere, the 
plan c being obtained by projection. 

With centres c and c, radius c/b', describe circles ; 
these are the projections of the required sphere. 

Examples. 1. A sphere 2'' diameter rests on the ground, with its 
centre C 1^" from the vertical plane. Determine a sphere, 
1 1" diameter, to touch the given one at a point P, whose plan 
p is 2^" from xy, and i" to the right of c. 

2. A sphere 2|" diameter, centre C, rests on the ground. 
Determine a sphere which shall rest on the ground, and 
touch the given sphere at a point whose elevation is 1" 
above xy and Jf" to the right of c' . 



XV 



SURFACES IN CONTACT 



371 




372 PRACTICAL SOLID GEOMETRY chap. 

306. Problem. To determine the projections of a 
sphere of given radius which shall touch (externally) a 
given vertical cone, at a given point P on its surface. 

Through p' draw al> parallel to xy, and draw b's at 
right angles to v'd' to intersect the elevation of the axis of the 
cone in /. If a circle (not shown) were described with s 
as centre and s'b' as radius, it would be the elevation of a 
sphere inscribed in the cone, and touching it in the 
circle whose elevation is db' . 

The required sphere must touch this sphere at P, 
hence its projection may be found as in Prob. 304, and is 
shown in the figure. 

307. Problem. To determine the projections of a 
sphere of given radius which shall touch a given inclined 
cylinder, axis DE, at a point whose plan p is given. 

Determine the elevation of /'(see Prob. 278, note 1), and 
also the projections of the sphere, centre S, which passes 
through P.a.nd is inscribed in the cylinder. 

The required sphere must touch this sphere at P; its 
projections may therefore be found as in Prob. 304, and 
the construction is shown in the figure. 

308. Problem. To determine the projections of a 
sphere which shall rest on the ground and touch a given 
inclined cone at a given point P (no figure). 

Obtain the projections of the sphere which is inscribed 
in the given cone and passes through P (see Prob. 278, 
note 1). The required sphere must touch this sphere at 
P, and its projections may be found as in Prob. 304. 

Examples. 1. A cone rests with its base on the ground ; diameter 
of base 2", height 3". Draw the plan and elevation of a 
sphere, i#" diameter, which shall touch the given cone at a 
point 2" from the vertex ; the elevation of this point being -j?" 
to the right of that of the axis of the cone. 

2. The plan and elevation of the axis of a cylinder, 2" 
diameter, make angles of 45 and 30 with xy. Determine 
a sphere, 1^" diameter, which shall touch the given cylinder 
at a point whose plan is " from the plan of the axis of the 
cylinder. 



XV 



SURFACES I\ CONTACT 



373 



\W 




374 PRACTICAL SOLID GEOMETRY chap. 

309. Problem. The plans of a sphere, centre C, and of 
a line AB which touches the sphere are given, the indices 
of c and a being attached ; determine and index the plan 
of the point of contact. 

Regard ab as the plan of a vertical plane ; the plane 
will intersect the sphere in a circle, diameter de, to which 
AB must be a tangent. 

Take xy parallel to ab, and project the elevations of 
A and the circle centre O, making md = 5 units. From a 
draw the tangent a'p' ; then this is the elevation of AB, and 
P is the required point of contact. Measure p'n and pro- 
ject and index/ accordingly. 

310. Problem. The circles, centres v and c, are the 
plans of a cone and sphere which rest on the ground, 
the height of V being given. Determine the plan of a 
cylinder of given radius which rests with a generator on 
the ground and touches the given cone and sphere. 

On xy, parallel to vc, draw the elevations of cone and 
sphere. 

Describe a circle, centre a^, radius equal to that of the 
required cylinder, to touch xy and the elevation of the 
sphere. Describe an equal circle to touch xy and e'v . 

These circles may be regarded as end elevations of the 
cylinder, and the plan of the axis of the required cylinder 
is distant from c and v lengths equal to aJn and b^m. 
Hence, with centres c and v, radii a/n and ml\, describe 
arcs and draw ab tangential to these arcs. This is the plan 
of the axis of the required cylinder, and the outline of the 
plan can now be at once drawn. 

Examples. 1. A sphere, 2" diameter, centre C, rests on the ground. 
A line a_.b, J" from c, is the plan of a tangent to the sphere at 
B ; _ 5 6=2j". Determine and index the plan b. Unit o. 1". 
2. A cone rests with its hase on the ground, diameter of base if", 
height 2^". A sphere ij" diameter also rests on the ground, 
its centre being i^" from the axis of the cone. Determine a 
plan of a cylinder, 1" diameter, which rests with a generator on 
the ground and touches the given cone and sphere, and show 
the points of contact. 



XV 



SURFACES IN CONTACT 



375 




376 PRACTICAL SOLID GEOMETRY chai\ 

311. Problem. To determine the projections of two 
cones which rest on the horizontal plane in line contact 
with one another, the vertical angles being e and 9. 

One axis VZ is taken parallel to the vertical plane. 

Since the cones must touch along a common generator, 
and also lie on the ground, their vertices must coincide. 

Draw the isosceles triangles v't'f and v't'g-^, with the 
vertical angles at v equal to 6 and <. These are the 
elevations of the two cones in line contact, with their axes 
parallel to the vertical plane. 

Take any point s in the axis v'z and draw s'a'^ per- 
pendicular to v't' . Then ^ and C x are the centres of two 
inscribed spheres touching each other at A v 

Project vsz parallel to xy ; with centre s, radius s'a' v 
describe a circle ; the tangents from v to this circle form 
the plan of the larger cone. 

Conceive the cone V\V to roll over the other until it 
lies on the ground ; its projections when in this position 
must now be found. 

Draw 1m parallel to, and distant C-[a^ from xy ; also 
draw c/c perpendicular to v'z , meeting lin in c . With 
centre c describe a circle to touch xy, and from v draw 
the upper tangent. The elevation of the smaller cone 
is thus found. 

With centre s', radius s'c^, describe a circle cutting 
hn in e ; draw s'r at right angles to hn. With centre s, 
radius re, describe an arc to cut a projector from c in 
c ; then c is the plan of C. The plan of the smaller cone 
is completed by drawing tangents from v to the plan of 
the sphere, centre C. 

To explain the construction. As the cone VIV rolls over the 
other, the triangle SCV turns about SV until C comes into the plane 
LM; c^c is the edge elevation of the circular path of C. During this 
motion C remains on the surface of a sphere, centre S, radius s'c^ ; 
hence the plan c is found. 

Note. Draw a-[ci parallel to c^c' , and project a on sc. Then A 
is the point of contact of the spheres, and VA the line of contact of 
the cones. 



XV 



SURFACES J\ CONTACT 



377 




Example. Determine the projections of two cones which rest on 
the horizontal plane in line contact with one another, the ver- 
tical angles being 50 and 25 respectively. 



378 PRACTICAL SOLID GEOMETRY chap. 

312. Problem. The scales of slope of two planes are 
given, to determine the plan of a cone, vertical angle 
e, which lies between the planes so as to touch them. 

The figure to the left shows an elevation of the cone 
with a sphere of any radius z inscribed in it. 

Conceive the cone lying between, and in contact with, 
the given planes. The vertex Fmust be at some point on 
their intersection, and C, the centre of the inscribed sphere, 
will be on the intersection of two planes respectively 
parallel to the given planes and distant z therefrom ; C will 
also be situated on the surface of a sphere, centre V, and 
radius V-[c^. 

Determine r's and lik\ edge views of the given planes, 
and draw lm' and f'g respectively parallel to and distant 
z from them. 

Determine a l> 10 , the plan of the intersection of the given 
planes, and also d Q e 1Q , the plan of the intersection of the 
second pair of planes (see Prob. 244). 

In a & 1Q select any point (in AB\ say z> 5 , as the plan of 
the vertex, and with v b as centre and radius v-[ c^ describe 
a circle ; this is the plan of a sphere on the surface of which 
C must be situated. But C is also in DE. 

Therefore by a construction similar to that in Prob. 309 
determine the intersection of DE and this sphere ; the plan 
of one point of intersection is c n - s . With centre c and 
radius z describe a circle and draw the tangents to it from 
v 5 ; the required plan of the cone is thus determined. 

Examples. 1. Draw oc and od including 120". On oc take 
or = |" and os = 2^". On od take oc = h", oc 2". Regard r Q s 15 
and f-5e 10 as the scales of slope of two planes. Determine the 
plan of a cone, vertical angle 45, which lies between the planes 
so as to touch them. Unit 0. 1". 

2. Determine the plan and elevation of a cone of indefinite length, 

vertical angle 6o, to which the planes of projection are both 
tangential. Show the lines of contact. 

3. The traces vt, th of a plane make 40 and 60 with xy. Find 

a right-angled cone which touches this plane and the ground. 



XV 



SURFACES IN CONTACT 



379 




General Examples. 1. Three spheres of diameters 2", ih", and 
1" rest on the ground in mutual contact, antl a fourth sphere, |" 
diameter, rests on the three. Draw the plan of the group. 
Draw also the plan of the triangular pyramid which circum- 
scribes the spheres, showing the three points of contact on 
each face of the pyramid. 

2. A line AB, 3" long, has its ends in the planes of projection, and 

is inclined at 6o to A'Kand 40 to the ground. Determine 
a cylinder having XY as its axis, which shall touch the 
line AB. 

3. Determine a sphere of 2" radius which shall have its centre in 

A'Kand touch the line AB of Ex. 2. 

4. A sphere, 3" diameter, has its centre C in XY; a. second sphere, 

ivr" diameter, centre S, touches both planes of projection ; 
SC=2h"- Determine a sphere which shall rest on the ground 
and touch the cone circumscribing the given spheres at a point 
distant f " from both planes of projection. 



380 PRACTICAL SOLID GEOMETRY chap. 

313. Problem. A given cone, axis VZ, lies on the 
ground ; to determine the projections of a cylinder of 
given radius which shall touch the cone externally at a 
point whose plan p is given, the direction, mn, of the 
plan of the axis of the cylinder being also given. 

Suppose the required cylinder in position touching the 
given cone at P, then 

(1) There will be two spheres inscribed one in each so 
as also to touch one another at P ; call their centres C 
and S. 

(2) The generators of cylinder and cone which pass 
through P must both lie in the tangent plane common to 
the cylinder and cone, or the inscribed spheres, at P. 

Determine, by Prob. 278, the elevation of P and the 
projections of the sphere, centre S, which passes through P 
and is inscribed in the cone. 

Also by Prob. 278 determine the projections of a 
sphere, centre C, which touches the given cone at P, 
and has a radius equal to that of the required cylinder. 

The required cylinder must circumscribe this sphere, 
hence its plan is obtained by drawing those tangents to 
the circle, centre c, which are parallel to mn. 

Draw vt perpendicular to sp. Then vt is the horizontal 
trace of the tangent plane referred to in (2) above. 

Draw ph parallel to mn, and project Ji ; join p'h' . 
Then p'h' is the elevation of a generator PH. Hence 
the elevation of the cylinder is obtained by drawing those 
tangents to the circle, centre c , which are parallel to p'h'. 

Note. If it were required that the cylinder should touch the 
cone along a generator, so as to have line instead of point contact, the 
tangents to the circle c, giving the plan, would be drawn parallel to/?'. 

Example. Draw two lines including an angle of 45 ; these form 
the outline of the plan of a cone which touches the ground along 
a generator. Determine the projections of a cylinder ij" 
diameter, the plan of whose axis makes 30 with the plan of the 
axis of the cone, which shall touch the cone at a point whose 
plan/ is 1" and \" respectively from the first two lines. 



XV 



SURFACES IN CONTACT 



38i 




General Examples. 1. A sphere 2\" diameter has its centre C 
in A'F. A point A is distant i" and 2" from the vertical and 
horizontal planes of projection and 3" from C. Determine a 
sphere, centre A, which shall touch the given sphere. 

2. A cone, vertex V, vertical angle 6o, has its axis along XY. A 

point A is distant 1" and i|" from the horizontal and vertical 
planes, and 2" from V. Determine a normal from A to the 
cone, and a sphere, centre A, which touches the cone. 

3. Draw the projections of any point A, and of any cone with its 

axis inclined to both planes of projection. Determine the two 
normals from A to the cone, and the two spheres, centre A, 
which touch the cone. 

4. A cylinder 2" diameter and a cone, vertical angle 60, lie on the 

ground in contact with their axes at right angles. Draw their 
plans and determine a sphere of \\" radius which shall rest on 
the ground and touch both. 



382 PRACTICAL SOLID GEOMETRY chap. 

314. Miscellaneous Examples. 

1. A sphere of 2^" diameter touches both planes of projection. A 

second sphere, diameter i|", touches the first sphere, and has 
its centre in the ground line. Draw the projections of the two 
spheres. (1889) 

2. Two spheres, diameters 2.25" and 1", rest on the horizontal 

plane touching each other. Draw the plan of the complete 
locus of the centre of a sphere of 1.4" diameter, touching both 
spheres. (1888) 

3. A right circular cone, the vertical angle of which is 35 , rests 

with a generator in the horizontal plane. A sphere of i|" 
radius also rests on the horizontal plane, and touches the cone 
in a point 2|" from the vertex. Draw the plan of the two 
solids. (1892) 

*4. Draw a sphere, I ' radius, resting on the horizontal plane, and 
touching the two given planes. Determine the points of con- 
tact. Unito.i". (1885) 

*5. Two planes are given by their traces aob, cod. Draw the pro- 
jections of a sphere 2" in diameter, touching the given planes, 
and having its centre in the horizontal plane. (1887) 

*6. Draw the plan of any sphere such that the given line AB is 
tangent to it, and that the centre of the sphere is in the line 
CD. Unit = 0.1". (1894) 

*7. The plans of a right cylinder and of a sphere are given. A right 
cone, diameter of base 2^-", height 3^", stands on the horizontal 
plane and touches both cylinder and sphere. Draw its plan 
and show the points of contact. (1886) 

*8. Determine the centre and radius of a sphere to which the two 
given lines AB, CD shall be tangent. Unit = o.i". (1881) 

*9. A plane is given by its scale of slope, and a line AB by its 
figured plan. Determine the centre of a sphere of i|-" radius, 
touching the given plane and line, and resting on the horizontal 
plane. Unit o. 1". (1890) 

*10. A right cone is lying on its side upon the horizontal plane, b is 
its vertex, and ab is the plan of its axis, which is inclined at 
25 . The point c is the plan of the centre of a sphere, which 
also rests upon the horizontal plane, and which is in contact 
with the cone. Complete the plan. (1882) 



XV 



SURFACES IN CONTACT 



383 



Co/iy /he fiaures double siz&. 



10 20 30 40 

I I I h 





-5 

e 




-s 

-e 



9 




CL 



Ik 



c 



15 



CHAPTER XVI 

INTERSECTIONS OF SURFACES, OR INTERPENETRATIONS 

OF SOLIDS 

315. The general problem and its solution. When two 
solids penetrate each other, the nature of the line of inter- 
section of their surfaces depends upon that of each of the 
surfaces. Thus if both surfaces consist of a series of plane 
faces, they will intersect each other in a series of straight 
lines ; or if one or both of the surfaces be curved, the 
intersection will consist of one or more curves ; these curves 
may be plane, but generally are tortuous curves ; that is, 
such as cannot be contained by a plane. 

One solid may completely penetrate the other, entering 
at a closed curve or zigzag line where their surfaces inter- 
sect, and emerging at a second closed curve or gauche 
polygon. Or the interpenetration may be only partial, in 
which case the line of intersection generally consists of 
only one closed figure. For the case intermediate between 
these two, the surfaces of the solids touch in one or more 
points ; here we must expect to find a node at a point of 
contact, that is, two branches of the curve of intersection 
may cross at the point. 

Method of sections. The method most commonly 
adopted for determining points on the intersection of the 
surfaces of two solids, or of any two surfaces, is that known 
as the method of sections. The solids are supposed to be 
cut by a series of section surfaces, generally plane, but 
sometimes spherical or otherwise curved. The shapes of 



chap, xvi INTERSECTIONS OF SURFACES 385 

the sections are drawn in plan and elevation, and the series 
of points where these intersect are thus determined. These 
points are common to the surfaces of the two solids ; that 
is, are points in the required intersection. The section 
surfaces are chosen, if possible, so that the projections of 
the sections shall always be straight lines or circles both in 
plan and elevation, these being the only two forms that can 
be drawn without trouble. 

It would at first appear that the greater the number of 
sections taken, and hence of points determined, the greater 
would be the accuracy with which the curves of intersection 
could be drawn. This, however, would only be true if each 
point could be located with absolute accuracy, which is not 
possible. So the greater the number of points the more 
marked will be the result of any slight error in determining 
any one. To obtain the best results, comparatively few 
section surfaces should be taken, but their positions 
should be very carefully and judiciously chosen. 

In almost all cases there will be certain important . 
special points on the line of intersection whose projections 
ought to be found. Such, for instance, are all points which, 
in the projections, fall on the outlines of the figures. The 
projected curve of intersection generally touches the outline 
at such points, and these points often separate a visible from 
an invisible portion of the intersection. The general 
method of procedure should be first to select those section 
surfaces which give us the important special points. These 
may be sufficient to enable us to plot the whole curve. 
But should there be any wide intervals, one or two extra 
sections may be taken to fill up the gaps. 

In the. problems and figures which follow, the positions 
of the important sections are generally indicated. A 
section is projected in detail, and the corresponding points 
of the curve of intersection determined. The projections 
of the other sections are omitted to avoid confusing the 
figures, though these may have had to be drawn to enable 
us to give the complete intersections. 

2 c 



386 PRACTICAL SOLID GEOMETRY chap. 

316. Problem. To determine the plan and elevation 
of the section of the given sphere, centre 0, by the given 
oblique plane LMN. 

Here one of the two intersecting surfaces is a plane. 

In determining points on the curves, it is convenient to 
employ a series of sections by horizontal planes, for these 
project into straight lines and circles in plan, and of course 
into straight lines in elevation. 

One such section plane is drawn in edge elevation at p'q . 
It cuts the sphere in a circle of diameter s x s v and the plan 
of this circle is drawn with centre o. And it cuts the plane 
LMNrn. the line whose plan is pq, determined by projecting 
from/' to/ and then drawing/^ parallel to mn. 

The plans of the sections of sphere and plane intersect 
in s, s ; and the elevations /, / are found by projecting 
from s, s on to p'q' . 

The points S, S are common to the section plane PQ, 
the plane LMN, and the sphere, and are therefore points 
on the required intersections of the two latter. 

This illustrates the method of determining points by 
any section plane ; we must now select those sections which 
give rise to the most important points on the curves. 

First as to the upper and lower limits of the curve. 
Draw ok perpendicular to mn; let this be the plan of a line 
in the plane LAIN; the line will evidently intersect the 
required section in its highest and lowest points. To 
determine the levels of these we have drawn the elevation 
A x A x of the line, supposing it to have been turned into a 
position parallel to the vertical plane about a vertical axis 
through O. The points A v A v where this elevation cuts 
the elevation of the sphere, give the highest and lowest 
levels for the section planes. The two sections at these 
levels give the two points A, A ; at a', a the required 
curve is horizontal ; at a, a the tangents are parallel to mn. 

Next as to the points on the outlines. A horizontal 
plane through the centre of the sphere will cut the latter 
in a great circle which projects into the outline in plan ; 



XVI 



INTERSECTIONS OF SURFACES 



3*7 




this section gives the two points B, B on the curve, whose 
plans b, b are on the outline plan of the sphere. Observe 
that the plan of the curve touches the outline at the points 
b, b where the two meet ; note also that the points b, b 
separate the full and dotted parts of the curve, which repre- 
sent the visible and hidden portions. 

To obtain the points on the outline in elevation, we 
must take a section which projects into this outline. This 
is given by a plane through O parallel to the vertical plane. 
The elevation of the line where this plane cuts the plane 



388 PRACTICAL SOLID GEOMETRY chap. 

LMJY'is shown ; it intersects the outline in c, c . Thus the 
points C, C are given by this vertical section plane. 

Again observe that the curve and outline touch at /, /, 
and that the curve changes from a full to a dotted one or 
vice versa when it passes a point on the outline. 

The eight points S, A, B, C thus determined are almost 
sufficient to enable the curves to be plotted, especially as the 
latter are ellipses, to the form of which the eye is well 
accustomed. Otherwise one or two intermediate section 
planes will give the required additional points. 

Examples. 1. Copy Fig. 316 double size, and work the problem. 

2. A cone stands upright on the ground, draw the projections of its 

section by an oblique plane. Diameter of base 2.2"; height 
2.6". Horizontal trace of plane touches base, and both traces 
make 50 with xy. 

3. A square pyramid stands upright on the ground, obtain the 

projections of its section by an oblique plane. Height 2.5", 
side of base 1.8", making 30 with xy. Horizontal trace of 
plane 1.2" from plan of axis, both traces making 45 with xy. 

Hint. Take section planes which contain two long edges 
of the pyramid, either opposite or adjacent edges, or both, 
whichever give the best-conditioned constructions. 

317. Problem. To determine the projections of the 
line common to the surfaces of two cylinders, the 
axes AB and CD of which intersect at right angles, 
CD being vertical. 

Horizontal section planes will be convenient. 

Describe the semicircle on 00 as diameter, and divide its 
circumference into six equal parts. 

Consider the section plane whose elevation //;/ passes 
through 2. Set off l>2, 1)2 in plan, each equal to m.2 in 
elevation, and through the points 2 thus obtained draw 
lines parallel to ab. These lines form the plan of the section 
of the horizontal cylinder, while that of the vertical cylinder 
is the circle, centre c. The two sections intersect in plan 
in the four points marked 2, which must therefore be on 
the required plan of the line of intersection. Draw projectors 



XVI 



INTERSECTIONS OK SURFACES 



389 





from the points in plan to meet bn in 2', 2', which points 
are on the elevation of the line of intersection. 

Repeat this construction for the planes through the 
other points of division of the semicircle and draw curves 
through the points so determined, obtaining two curves for 
the elevation as shown, while the plan coincides with portions 
of the circle, centre c. 

Note. If the vertical cylinder had been the smaller one, section 
planes parallel to the vertical plane would have been preferable. 



39Q PRACTICAL SOLID GEOMETRY chap. 

318. Problem. To determine the projections of the 
curve of intersection of a given cone and cylinder, 
the axes of which intersect each other at right angles, 
that of the cone being vertical. 

Three Cases are shown. In (i) the cylinder completely 
penetrates the cone; in (2) the cylinder and cone circum- 
scribe the same sphere ; and in (3) the cone completely 
penetrates the cylinder. 

Case (1). The projections of the cone and cylinder 
are given in the figure, the diameter of the cylinder being 
equal to that of a circle described so as to fall within the 
elevation of the cone. This circle may be regarded as an 
end projection of the cylinder. 

Horizontal section planes are convenient, since the 
sections of the cone are circles and those of the cylinder 
pairs of parallel straight lines in plan. In elevation the 
sections are overlapping straight lines. 

On g'Ji describe a semicircle and divide its circum- 
ference into six equal parts. 

Through any of the points of division, say i', draw //// 
parallel to xy ; this is the edge elevation of a horizontal 
section plane. 

With v as centre and radius on describe a circle ; this is 
the plan of the section of the cone by the plane LM. 

Set off b\ =mi' on each side of b as shown, and draw 
lines through the points 1 parallel to ab ; these lines form 
the plan of the section of the cylinder by the plane LM. 

Hence the points in which these sections intersect are 
points on the required plan, their elevations being found by 
projectors as shown. 

Repeat this construction for the planes through^', i', 2 
. . . //, and draw a curve through the points thus found. 

The planes through 4, 5', H give rise to points which in 
plan are hidden by the cylinder, and the plane through 3' 
will divide the dotted and full portions of the curves in 
plan. The curve is seen to touch the outline of the cylinder 
in plan at these points. 



XVI 



INTERSECTIONS OF SURFACES 



391 




Examples on Problem 317. 

1. Determine the interpenetration of two cylinders, diameters lY 

and 2" ; axes intersecting at right angles, and parallel to vertical 
plane ; smaller cylinder horizontal. 

2. Work Ex. 1 (a) when the axis of the smaller cylinder is hori- 

zontal, but makes 30 with the vertical plane ; (I) when the axis 
of the smaller cylinder is inclined at 30 to the horizontal plane. 

3. A semi-cylinder, 4V' diameter, rests with its rectangular face 

on the ground. A cylinder, 2j" diameter, rests with a generator 
on the ground, at right angles to the axis of the semi-cylinder. 
Determine the plan of the curve in which the surfaces intersect. 



392 PRACTICAL SOLID GEOMETRY chap. 

Case (2). The diameter of the cylinder is taken equal 
to that of a circle which touches v'd' and v'e'. 

Repeat the construction of Case (1). If this be done 
accurately, the elevation will be found to be two intersecting 
lines. The intersection of the surfaces therefore consists of 
two plane sections of either surface, that is, of two ellipses. 
These ellipses project as ellipses in plan. In this case the 
cone and cylinder circumscribe the same sphere. See 
Arts. 228 and 273, and Theorems 23 to 48, Appendix II. 

An important section plane is that whose elevation 
passes through f, the point where v'e touches the circle; 
this section gives the points where the ellipses intersect in 
plan and elevation. Observe that at these points the 
surfaces touch one another, and two branches of their 
intersection cross one another at each point of contact. 

Case (3). The diameter of the cylinder is taken equal 
to that of a circle described so as to intersect the lines v'e, 
v'd', in points such as r, s. 

As important section planes take those whose elevations 
pass through g and r. Another plane between these will 
be sufficient to determine the upper curve of intersection. 

For the lower portion select the two important section 
planes whose elevations pass through s' and //, and in 
addition one or two planes between them. 

Observe that the lower curve is hidden by the cylinder 
in plan. 

Examples. 1. Draw an isosceles triangle abc, with the base 
fc=3i", and altitude ad=T, \ this is the elevation of a cone, 
vertex A, with its base on the ground. On da take de=i^". 
With c as centre describe a circle to touch ah and ac. Also 
describe two circles concentric with this, but having radii T V' 
less and T V' greater. Each circle is the end elevation of a 
cylinder. In each case determine the plan and elevation of 
the curve of intersection of the two surfaces, when the cylinder 
is turned with its axis parallel to the vertical plane. 

2. In Ex. 1 describe a semicircle on xy with centre c and radius 
2#" ; this is the end elevation of a semi-cylinder ; draw the 
plan of the curve in which the cone and semi-cylinder intersect. 



XVI 



INTERSECTIONS OF SURFACES 



393 



I 

a' 




T?pir l 

'\ ry :,Xr' v ^ 
/ Pi c TNl ' N * 

' / ' 1 ' >v^ 

/ \ / \ v 

v *. ''* Nl 


::\2' 

\ 

i 
1 
/ 


V 


d'\ 


! \e 


y 




394 PRACTICAL SOLID GEOMETRY chap. 

319. Problem. To determine the projections of the 
line of intersection of a given cylinder and cone, the axes 
of which are both vertical. 

Horizontal section planes are convenient. 

Consider a plane whose elevation is tin. With centre 
v and radius // describe a circle ; this is the plan of the 
section of the cone. The points/,/ in which this circle 
intersects the plan of the cylinder are points on the 
required plan ; the elevations p', p' are obtained by pro- 
jection. 

Sections which give important points. Draw vd through 
v and a, intersecting the circle, centre a, in d and c ; then 
D and C are the highest and lowest points on the line of 
intersection. Also draw ni through a parallel to xy ; then 
/', // are on the elevation of the outline of the cylinder. 
Hence planes should be taken which cut the cone in circles 
whose radii are equal to vd, vc, vn, and vi. Two or three 
intermediate planes may be required. 

Note. Vertical section planes containing the axis of the cone 
would be very convenient to take in working this problem ; the 
important points D, C, JV, I might be readily determined in this 
manner. Such planes would cut both surfaces in straight lines. 

320. Problem. To determine the interpenetration of 
a given vertical cylinder, axis AA, and a given sphere, 
centre S. 

Take vertical section planes parallel to XY. Consider 
one such plane of which /;;/ is the plan. 

Draw the circle with centre s and radius om; this is the 
elevation of the section of the sphere. The section of the 
cylinder is a pair of straight lines, one of which has c for 
its plan and c'c for its elevation. The line and circle 
intersect as shown in two points p',p\ which are on the 
required elevation, p being the plan of the points P. 

Important section planes are those whose plans pass 
through e, a, s, and f. One ' or two planes may be taken 
in addition to these. 

It will be observed that the required plan is the arc erf. 



XVI 



INTERSECTIONS OF SURFACES 



395 




Example on Problem 319. Describe two circles with radii i|" 
and ", their centres c and v being .85" apart ; these are the 
plans of a cylinder and cone standing upright on the ground ; 
height of cone 3". Determine the elevation of their curve of 
intersection on an xy making 35 with cv. 

Obtain the developments of the cone and cylinder, in each 
case showing the curve of intersection. 



396 PRACTICAL SOLID GEOMETRY chap. 

321. Problem. To determine the projections of the 
line of intersection of a given sphere, centre S, and a 
given vertical triangular prism. 

Either vertical or horizontal section planes may be 
employed, but the former are preferable in determining the 
important points. 

Consider the plane whose plan is 1m. With / as centre 
describe a circle, the diameter of which is ab ; this is the 
elevation of the section of the sphere. 

Draw the lines cc, ad' by projecting from c and d ; 
these form the elevation of the section of the prism. 

The intersections P of these two sections are points on 
the required curve. 

The important sections. Draw se, s/i, so respectively 
perpendicular to the sides of the triangle uzw, and take 
section planes through e, o, n ; they give the highest and 
lowest points on the three portions of the intersection of 
the sphere and prism. Take section planes through u, 
s, g } h, s ; the plane through s will give the points on the 
outline of the sphere in elevation. 

The curves of intersection are ellipses. 

Examples on Problems 320 and 321. 

1. Describe two circles with radii i-J" and i", the distance between 

their centres being |" ; these are the plans of a cylinder and 
sphere which intersect. Determine the elevation of the curve 
of intersection on an xy which makes 30 with the line joining 
the centres of the plans. 

2. Draw an equilateral triangle abv 3 V side ; bisect av in c. With 

centre c describe a circle to touch ab and av ; this figure is the 
plan of a cone, vertex V, and a sphere, the axis of the cone 
. being horizontal, and the centre of the sphere on the surface of 
the cone. Determine the plan of the curve of intersection 
and the elevation on an xy parallel to ab. 

3. Draw a triangle abc with ab = 2.7$", ^=3.2", ac = 2.f. Take 

a point s inside abc distant 1. 1" from ac and .9" from be ; with 
s as centre describe a circle 3.5" diameter. The circle and 
triangle are the plans of a sphere and prism. Determine the 
elevation of their curve of intersection on an xy making 20 
with ab. 



XVI 



INTERSECTIONS OF SURFACES 



397 





398 PRACTICAL SOLID GEOMETRY chap. 

322. Problem. To determine the plan and an elevation 
of the line of intersection of a pyramid and prism, the 
indexed plans of which are given ; the pyramid rests with 
its base on the horizontal plane, and the edges of the prism 
are horizontal. Unit o. i". 

Draw an elevation of the solids as seen when looking in 
a direction parallel to the long edges of the prism ; that 
is, take xy perpendicular to the plans of these horizontal 
edges. 

In this example the intersection will consist of a series 
of straight lines, which may be obtained by determining 
(i) the points in which the edges of the prism intersect 
the faces of the pyramid; (2) the points in which the 
edges of the pyramid intersect the faces of the prism ; and 
then joining these points by straight lines in the proper 
sequence. 

Join v'e and produce to w ; then v'tv is the elevation 
of two lines, one on each of the faces A VB, A VD of the 
pyramid, if vw, vz are their plans. This may be regarded 
as a section by a plane containing J^and the horizontal edge 
E. Hence H and K are the points in which the edge E 
meets the faces A VB, A VD of the pyramid. 

In a similar manner it is found that the edge F 
meets the faces CVB, CVD in L and M, and the edge 
G meets the same two faces in yVand O. 

Next consider the edge VB of the pyramid. It inter- 
sects the faces EF and EG of the prism in points whose 
elevations are t' and u, the plans t and it being obtained at 
once by projection. In like manner the edge VD meets 
the same two faces in R and S. 

When joining the points thus found, each face of the 
prism may be taken in turn and its lines of intersection 
with the faces of the pyramid noted. 

Thus, taking the face EG, it will be seen that we must 
join hit, un, os, sk ; all of which are underneath the prism 
and therefore dotted. On the face GEwe get In, mo; and 
on the face EE we have ht, //, mr, rk. 



XVI 



INTERSECTIONS OE SURFACES 



399 




400 PRACTICAL SOLID GEOMETRY chap. 

323. Problem. The indexed plan of an irregular tetra- 
hedron is given, the circle representing a cylindrical hole 
bored vertically through the solid. Draw the elevation 
of the pyramid on a vertical plane parallel to AB. 

Take xy parallel to ab and draw the elevation. 

The hole passes through the face ABC and partly 
through the faces ABD, CBD. 

A series of vertical section planes passing through B 
will in this case be convenient. 

Consider the section plane whose plan is be ; it cuts the 
pyramid in a triangle BFE and the surface of the hole in 
two vertical lines whose plans are /// and n respectively. As 
obtained from the elevations of the lines these two sections 
intersect in T, U, M, and IV, which are therefore points on 
the required curve. 

Important Sections. Through o draw rs parallel to ab. 
Then the section planes whose plans pass through /' and 
each of the points r, s, d are important ; also the planes 
which touch the surface of the hole. In connection with 
the last two planes it should be observed that the lines in 
which they intersect the faces of the pyramid are tangential 
to the curve of intersection. This will appear on drawing 
the elevation, and the points of contact should be found. 

Examples on Problems 322 and 323. 

1. Draw a quadrilateral abed, having a^ = 3f", bc = 2", ac=^", 

ad=2", cd=2^". Take an inside point v i" from ab and i-J" 
from ad; join v to a, b, c, d. This is the plan of a pyramid 
with its base ABCD on the ground, and the vertex V 3" high. 
An equilateral triangular prism with its long edges horizontal and 
perpendicular to AC penetrates the pyramid. A side of the end 
of the prism is 1^" ; one face of the prism is horizontal, its 
centre being 2|" vertically below I'. Determine the plan of the 
intersection and the elevation on an xy parallel to ac. 

2. A quadrilateral a.^b.^c^d^, with the diagonals ac, bd, is the plan 

of an irregular pyramid; ab=$", bc ^', ac = 4^", cd=T,", 
and ad= 2^". A circle, diameter 2", with its centre on ac and 
touching cd, is the plan of a vertical cylindrical hole cut through 
the pyramid ; determine the elevation of the pyramid and hole 
on an xy parallel to cd. Unit o. 1". 



XVI 



INTERSECTIONS OE SURFACES 



401 




2 D 



402 PRACTICAL SOLID GEOMETRY chap. 

324. Problem. To determine the projections of the 
intersection of two surfaces of revolution given in eleva- 
tion, the axes of which intersect each other and are 
parallel to the vertical plane, one being vertical. 

The given surfaces are generated by the revolution of a 
circular arc about an axis in the plane of the arc ; in one 
case the axis, CD, intersects its arc ; and in the other the 
axis, AB, does not intersect it. We shall employ the 
method of spherical sections. 

Let the axes intersect in O, and with o as centre 
describe a circle as shown in the figure ; let this be the 
elevation of a spherical section surface which intersects 
each given surface of revolution in a pair of circles, each 
circle of one pair cutting one of the circles of the other 
pair. Thus the circles whose elevations are e'f and g'h' 
cut each other at points which have r for elevation. 

In like manner s is obtained from the other two circles. 

To obtain the plans of these points : with centre o 
describe a circle, the diameter of which is g'ti ; draw pro- 
jectors from r, s' to meet the circle in r, s. 

Repeat this construction for two or three other spherical 
sections, centre o. 

The plans of the surfaces and of the curves of intersec- 
tion are not shown in the figure. 

Since the outlines of the surface in elevation may be 
regarded as sections by a plane containing the two axes, it 
follows that M, N~, U, V are on the curve of intersection. 

Examples. 1. Determine the plan and elevation of the curve of 
intersection of two surfaces of revolution like those in the figure, 
whose axes intersect, the axis of one being vertical and that of 
the other inclined at 45 to the ground. The least cross- 
section of the one is ii" in diameter, and the greatest cross- 
section of the other is i4". The radius of the curved outline of 
each is 3^". 

2. The axes of a cylinder and a double cone of indefinite length in- 
tersect each other at a point ^" from the vertex of the latter. 
Vertical angle of cone 55 , axis vertical. Diameter of cylinder 
l" ; inclination of axis 45. Draw the plan and elevation of 
their curve of intersection. 



XVI 



INTERSECTIONS OF SURFACES 



403 



1 a 




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\ '' 




\ > f / 

C ' \w /.. } ._ / 

\ "> wC/r' ~^\ ' \ /// 
\ v <7"*\ !---^----^ 

\ C/ ' ' \ ' \ h 

\ i '"'' \ >' \//A 




Ai7 /r <f r /T\ 

' / x v / \ A > 

K "^"""' "1" c\'\ -- 




/ p \l> \ 




X 


J 



,-v 



V'4 



<? 



^ 



\r 



S'h' 



404 PRACTICAL SOLID GEOMETRY chap. 

325. Miscellaneous Examples. 

*1. Fig- () The plans of the edges of a four-faced right prism 
with a horizontal axis are given. The prism is penetrated by a 
vertical square prism 2|" high. Draw an elevation on a plane 
parallel to the axis of the horizontal prism, showing the invisible 
portions of the intersection by dotted lines. Unit = o. i". 

(1886) 

*2. Fig. (b). The elevation of a right cone resting on the horizontal 
plane is given. The hatched semicircle is the elevation of a 
hemi-cylindrical portion cut out of the cone. Draw the plan 
of the remaining portion of the cone. O893) 

*3. Fig. (c). The given pyramid on quadrilateral base ABCD in 
the horizontal plane, and vertex v. 27 , is penetrated by the given 
triangular prism, the edges of which are horizontal. Determine 
the intersection of the surfaces of the two solids. Unit = o. 1". 

(1890) 

*4. Fig. (</). The flat base of a hemisphere of i|" radius (centre O ) 
rests on the ground, as does also that of a pyramid abed, the 
height of its vertex {V) being 2^". Draw the plan and eleva- 
tion on xy of the intersection of the pyramid and hemisphere, 
and develop the faces BVC, DVC so as to show the develop- 
ment of the intersection. ( 1896) 

*5. Fig. (e). A vertical triangular slot is cut through a sphere. The 
plan of the sphere and slot is given. Draw an elevation of the 
sphere. (1879) 

*6. Fig. (_/"). It is required to fit a cylindrical steam dome on the 
top of a cylindrical boiler shell. A sketch with the required 
dimensions is given. Draw the elevation of the curve of inter- 
section of the dome and shell, and obtain the development of 
the dome. Scale ^ z . (1880) 

7. A horizontal cylindrical hole (diameter ij") is bored through a 

vertical cylinder (diameter 2|") ; the axis of the boring cylinder 
passes j" from the axis of the vertical cylinder and is inclined 
2 5 to the vertical plane. Draw the elevation of the vertical 
cylinder. ( 1S77) 

8. A cylinder and sphere of Ij" and l|" diameters respectively 

rest with their curved surfaces on the ground, the centre of 
the sphere being in the surface of the cylinder. Draw the 
plan of the curve of penetration and an elevation of the curve 
on a vertical plane making 45 with the axis of the cylinder. 

9. A cone, height 3", rests with its base on the ground, diameter of 

base 3". The axis of a cylinder 2" diameter is parallel to the 
vertical plane and inclined at 45 to the ground, and passes 
through the vertex of the cone. Determine the plan and eleva- 
tion of the curve of intersection of the surfaces. 



XVI 



INTERSECTIONS OE SUKEACES 



405 



Copy the figures double stye. 
8 26 o is 




CHAPTER XVII 



CAST SHADOWS 



326. Preliminary. It is a matter of common observa- 
tion that the rays of light which emanate from any source 
proceed in straight lines in all directions through space, 
except so far as they may be intercepted by opaque objects, 
or otherwise influenced. If a surface receive the rays, a 
portion of the surface will be deprived of light by the inter- 
position of the opaque body. The space devoid of rays 
between the body and the surface is the shadow of the 
body, and that part of the surface deprived of light is 
the shadow cast by the body on the surface, or the cast 
shadow. Thus a lunar eclipse occurs when the moon 
enters the region of the earth's shadow, and as we watch 
the phenomenon we see the earth's cast shadow on the face 
of the moon. 

The shadows we see around us are generally of a most 
complicated nature. Although there may be only one 
original source, yet all the surrounding objects on which 
the light falls give back some of the light, and thus become 
secondary sources of greater or less intensity and of varying 
size ; these reflected lights play a very important part in 
pictures. We give no account here of this maze of varying 
light and shade. Our task is comparatively simple ; we 
confine attention to one source of illumination, and this 
is further supposed to be so small that it may be treated 
as if it were a point. The corresponding shadows may 
be termed geometrical. The nearest approach to this in 



chap, xvn CAST SHADOWS 407 

nature is perhaps an electric arc light, the cast shadows 
from which are very sharply defined. When the sun is the 
source, the shadows are softened at the edges on account of 
the angular magnitude of the sun's disc. 

If the point source is near at hand we have divergent 
rays, but we generally simplify the problems still further, 
and assume that the source is so far away that all the rays 
are practically parallel to one another. 

327. Theorems relating to geometrical shadows. The 
opaque object which casts the shadow receives light from 
the source on one portion of its surface, the other portion 
being in shade. The line on the surface which divides 
the two portions is called the line of separation ; this line 
is evidently the locus of the point of contact of those 
rays which touch or graze the surface ; it is further evident 
that these bounding or extreme rays are those which define 
the outline of the shadow cast on any surface. We thus 
have 

Theorem 1. For any object which casts a shadow, the 
line of separation on its surface is the locus of the point of 
contact of the bounding rays ; and the shadow cast by the line 
of separation on any surface is the outline of the shadow cast 
by the object itself on the same surface. 

We may if we like regard the rays of light as projectors, 
in which case the shadow cast by any body on a plane 
becomes its outline projection on the plane. The projec- 
tion is parallel (and oblique) ox radial, according to whether 
the rays are parallel or divergent. 

Under another aspect the cast shadow may be looked 
upon as the section of a cylinder or cone, the latter terms 
being here used in their wider meanings (see Definitions 
24 and 27 of the Appendix); the bounding rays become 
the generators of the cylindrical or conical surface, and the 
line of separation is the directing curve. 

Thus theorems relating to cast shadows may often be at 
once deduced from familiar theorems of projection, or of 
conic sections, e.g. 



408 PRACTICAL SOLID GEOMETRY chap. 

Theorem 2. The shadow cast by a point on any surface 
is the trace on that surface of the ray through the point. 

Theorem 3. The shadow cast by a straight line on any 
plane is the straight line joining the shadows cast by its ends. 

Theorem 4. The shadow of a straight line on any surface 
coincides with the trace on that surface of a plane which con- 
tains the straight line and the source, for divergent rays, or 
which contains the straight line and is parallel to a ray, for 
parallel rays. 

Theorem 5. If a series of straight lines be parallel to one 
another, their shadoivs on any plane by parallel rays are also 
parallel to one another and of proportionate lengths. If the 
rays are divergent the shadoivs are also divergent. 

Theorem 6. If a plane figure be parallel to a plane, its 
shadoiv on the plane is equal and similar to the figure if the 
rays be parallel, and is similar if the rays be divergent. 

This last theorem follows from a property of the cylinder 
or cone, viz. that parallel sections of a cone (or pyramid) 
are similar figures, and of a cylinder (or prism) are similar 
and equal figures. Thus for parallel rays the shadow cast 
on the ground by a horizontal circle is a circle of the same 
size, the centre of the latter being the shadow cast by the 
centre of the former. 

In regard to the general method of working problems, 
in some cases it is necessary to determine the line of sepa- 
ration before we can obtain the cast shadow ; in others the 
shadow helps us to determine the line of separation ; in all 
cases the connection between the two should be constantly 
borne in mind. Tf the object have any corners, the 
shadows of these should be found. When a shadow is cast 
on two surfaces which intersect, such as the two planes of 
projection, those points on the outline of the shadow which 
fall on the intersection should be specially determined as 
important points. 

The following problems are confined to parallel rays. 
The direction of the latter may be defined by the projec- 
tions, or by the indexed plan, of any single ray. 



XVII 



CAST SHADOWS 



409 



W 







<? 



(b) 



328. Problem. To determine the shadow cast by a 
given point P, (a) on the horizontal plane, (b) on the 
vertical plane, having given the direction of the parallel 
rays in plan and elevation. 

(a) Draw the projections of a ray through P ; that is, 
through p draw a line parallel to the plan, and through p' a 
line parallel to the elevation of the given direction of the 
rays. Determine / , the horizontal trace of the ray ; then 
f> is the required shadow. 

(/?) Draw the projections of a ray through P and deter- 
mine its vertical trace p x ; then p x is the required shadow. 

Examples. 1. A point A is 2" in front of the vertical plane and 
1" above the ground. The plans and elevations of the rays 
make angles of 30 and 45 with xy. Determine the shadow 
of the point on the ground. 

2. Take the point A in Ex. I to be 1" from the vertical plane and 

2" above the ground ; find the shadow on the vertical plane. 

3. A point A is 2" above the ground and 1" from the vertical plane. 

(1) Determine whether the shadow is cast on the vertical or 
horizontal plane, the rays being inclined at 40 and their plans 
making 30 with xy. (2) The direction of the plan being 
unaltered, what must be the inclination of the ray if the shadow 
of A is on xy, and what if the shadow is |" below xy? 



410 PRACTICAL SOLID GEOMETRY chap. 

329. Problem. To determine the shadow cast by a 
given line AB, (a) on the horizontal plane, (b) on the 
vertical plane, (c) on both planes of projection, having 
given the direction of the parallel rays in plan and 
elevation. 

(a) By Prob. 328 determine a and b , the shadows cast 
by A and B on the horizontal plane ; then a b is the re- 
quired shadow of the line. 

(b) Find by Prob. 328 a l and b v the shadows cast hyA and 
B on the vertical plane ; then a x b x is the required shadow. 

{c) Determine a b , the shadow cast by AB on the hori- 
zontal plane, supposing the vertical plane to be transparent 
like glass ; also determine b v B's shadow on the vertical 
plane. Let a b meet xy in i ; join t' b r 

Then the broken line a / b x is the required shadow, a i 
being on the horizontal plane and shown in plan, while i b x 
is on the vertical plane and is shown in elevation. 

Examples. 1. -/ is a point 2" from each plane of projection ; B 
is 3" from the vertical plane and 1" above the ground ; AB is 
3" long. Determine (1) the shadow cast on the ground ; (2) the 
shadow cast on a plane parallel to the vertical plane and 1" in 
front of it ; (3) the shadow cast partly on the vertical plane 
and partly on the ground, and determine the point on AB 
whose shadow is in xy. The plan and elevation of a ray make 
30 and 6o with xy. 

2. a-n/'io is 2" long and is the indexed plan of a line AB ; c is 2" 

from b and 1" from a. The plan of a ray makes 60 with ab, 
while the inclination of a ray is 50 . If the shadow of C falls 
on the shadow of AB, determine the index of c. Unit = o. 1". 

3. The plan and elevation of a ray both make 45 with xy 

determine the inclination of the ray. Am. 35-2. 

4. A system of parallel rays make 45 with xy in plan, and their 

inclination is 45 ; find their direction in elevation. 

5. A triangle ABC has the point A on the ground, and its plan is 

an equilateral triangle abc of 2.5" side. The shadow of ABC 
on the ground is a right-angled isosceles triangle with a as 
vertex, the parallel rays having an inclination of 45, and their 
plan making 45 with ab and 15 with ac. Determine the 
shadow a^ r ; index the points b and c ; and find the true 
shape of ABC. 

Hint. Make use of Prob. 21 to draw the shadow. 



XVII 



CAST SHADOWS 



411 




Examples on Problems 330 and 331. 

1. A cube 2" edge rests with a face on the ground, one side of the 

base making 40 with xy, the nearer end of this side being 1" 
from the vertical plane. Determine the shadow on the planes 
of projections when the rays in plan and elevation make angles 
of 45^ with and are directed towards xy. Indicate the two 
portions of the line of separation which give rise to the two 
parts of the shadow, namely, those which are cast on the 
horizontal and vertical planes. 

2. Copy double size the figured plan of the tetrahedron on p. 199, 

and attach the same indices; unit o. 1". Find the shadow of 
the solid on the ground, without drawing an elevation, making 
use of the following theorem. The rays are inclined at 45, 
and in plan are parallel to ab. 

Theorem. For parallel rays, the line joining the plan of a 
point to its shadow on the ground is parallel to the plan of a ray, 
and of length proportional to the height (or index) of the point, 
this length being equal to the height when the inclination of the 
ray is 45 . 

3. An octahedron of 2" edge rests with a face on the ground. 

Draw its plan, and by the method of Ex. 2 determine its 
shadow on the ground, the rays being inclined at 35 , and in 
plan making 45" with a horizontal edge of the solid. 

4. Determine the shadow cast on both planes of projection by the 

pyramid in Ex. 23, page 201, the vertical plane being parallel 
to ad, the rays inclined at 45 and parallel, in plan, to av 
Indicate the line of separation and the portions of it which cast 
the shadows on the ground and vertical plane respectively. 



412 PRACTICAL SOLID GEOMETRY chap. 

330. Problem. A given cube rests with one face on the 
ground ; to determine its shadow on the latter from given 
parallel rays. Also to find the line of separation on the 
cube. 

The projections of the cube are shown in the figure. 

Consider the vertical edge BB. Obtain b , the shadow of 
the upper end B. Join bb ; then bb Q is the shadow of BB. 

I )raw cr and dd , each parallel and equal to bb ; join 
b c and c d . Then bb t d dab is the outline of the required 
shadow of the cube. 

It will be evident that the outline of the shadow is cast 
from the vertical edge BB, the upper horizontal edges BC, 
CD, the vertical edge DD, and the lower horizontal edges 
DA, AB ; hence these edges constitute the line of separa- 
tion, the vertical faces BC, CD and the base of the cube 
being in shade. 

331. Problem. The indexed plan of an irregular pyra- 
mid, with its base ABCD resting on the ground, is given, 
and also r n s 7 , the indexed plan of a parallel ray. To 
determine the shadow of the pyramid on the ground, and 
on the vertical plane the plan of which is lm. Also to 
indicate the line of separation on the solid. Unit 01". 

On //// as a ground line determine the elevations of BS 
and V. Obtain v Q and v v the horizontal and vertical 
traces of a ray through V, and join v Q b, v d, meeting /;;/ in 
/o and e . 

The lines v b, v d, together with ab, ad, would form the 
outline of the shadow of the pyramid on the ground if the 
vertical plane LM were transparent or were removed. 

Join e v v f f) V 1 ; then b/ e dab is the outline (in plan) of 
that part of the shadow which falls on the ground, and 
f v x e Q is the outline (in elevation) of the remainder of the 
shadow, which falls on the given vertical plane. 

Since the outline of the shadow is cast from the edges 
VB, BA, AD, and D V, these edges constitute the line of 
separation. 



XVII 



CAST SHADOWS 



413 







414 PRACTICAL SOLID GEOMETRY chap. 

332. Problem. A given circle, centre C, is parallel to 
the vertical plane, to determine the shadow cast by it on 
the planes of projection, having given the rays. 

We must first find the shadow cast on the vertical plane, 
on the supposition that the horizontal plane is transparent. 
That is, determine c v the shadow cast by C on the vertical 
plane, and with c x as centre describe a circular arc with a 
radius equal to that of the given circle ; this arc meets xy 
in d and e , and is that portion of the required shadow 
which falls on the vertical plane. See Theorem 6, Art. 327. 

Through d draw d d' parallel to the given elevation of 
a ray, and draw d'e parallel to xy. It will be obvious that 
the arc DSE is that which casts the shadow on the vertical 
plane ; the remaining arc DRE throws its shadow on the 
horizontal plane. 

To determine the latter shadow, take any point N on 
the circle ; determine n ; draw n Q m Q parallel to xy, making 
n m = u'm'; then // , m are two points on the required 
shadow. Repeat this construction for one or two other 
points on arc RJYD, and draw the elliptical arc d ce through 
the points so obtained. This will be the required shadow 
on the horizontal plane. 

Note that at the point of contact of the circle and 
ground, the tangents to the shadow and circle coincide. 

An important ray is the one which touches the circle in 
elevation at say t' (not shown). The corresponding 
projector for the shadow touches the latter at / . 

333. Problem. To determine the shadow cast by a 
given sphere on the ground, the given rays being parallel. 
To find also -the line of separation on the sphere. 

Suppose the sphere to be circumscribed by a cylinder, 
the axis of which is parallel to the given rays, then the 
horizontal trace of the cylinder will be the required shadow, 
which may therefore be found as in Prob. 277. 

The line of separation is the circle of contact of the 
cylinder and sphere. 



XV II 



CAST SHADOWS 



415 




Examples. 1. A circle 2" in diameter has its plane parallel to 
the vertical plane and 1" therefrom, its centre being 1^" above 
the ground. Determine the shadow cast on the planes of 
projection, the rays in plan and elevation making angles of 
35 and 40 respectively with xy. 

Determine the projections of that chord of the circle whose 
shadow coincides with xy. 

2. Suppose that the circle in Ex. i has its plane horizontal, the 

centre being 2" above the ground and ii" from the vertical 
plane. Determine the shadow cast from it on the planes of 
projection. 

3. A sphere 2 : ' radius has its centre 2" from each plane of projec- 

tion. Determine the shadow on the ground if the rays on 
plan and elevation make angles of 30 and 45 with xy. 

4. Determine the shadow of the sphere in Ex. 3 cast on a plane 

inclined at 30 , and I A" from the centre of the sphere, the rays 
being parallel to the vertical plane, and inclined at 40 to the 
ground and UO to the inclined plane. 

5. In Ex. 4 what should be the inclination of the plane, so that 

the plan of the shadow is a circle, the rays being unchanged ? 



416 PRACTICAL SOLID GEOMETRY chap. 

334. Problem. To determine the shadow cast on the 
horizontal plane, (a) by a given cone resting with its base 
on the ground ; (b) by a given cone, axis vertical, with 
its vertex on the ground ; and (c) by a given cylinder 
resting with its base on the ground. Also to show the 
line of separation in each case. 

(a) Obtain v Q , the shadow of the vertex on the ground ; 
draw the tangents v Q f , 7> r Q ; then these tangents together 
with the arc / r form the outline of the required shadow. 

Draw the radii vt Q , vr Q at right angles to the tangents ; 
then the generators VJ\ VR and the arc TRN form the 
line of separation on the cone, the portion VTNR V being 
illuminated, while the remaining surface, including the base, 
is in shade. 

(b) Determine c , the shadow of the centre of the base 
of the cone, and with c as centre describe a circle with a 
radius equal to that of the base ; draw the tangents z// , vr Q . 
These tangents together with the arc t^n Q r Q form the outline 
of the shadow on the ground. 

Draw the radii cJ w c Q r Q perpendicular to the tangents ; 
and draw vt, vr respectively parallel to these radii ; then 
VT, VR and the arc TNR constitute the line of separa- 
tion, and the portion VTNRV of the conical surface is in 
shade. 

(c) Determine o Q , the shadow of the centre of the upper 
end, and with o as centre describe a circle with a radius 
equal to that of the cylinder. Draw the tangents tt {) , rr Q ; 
then these tangents together with the semicircles rmt, r Q n Q t Q 
form the outline of the required shadow. 

Draw the diameter rt at right angles to oo Q , then the 
generators RR, TT, the upper semicircle RNT, and the 
lower semicircle RA/T form the line of separation ; one-half 
of the cylinder, including the base, is in shade. 

Note. Observe that the shadows z' f , ~'(/o m ( a ) an( l (^), and 
rr , tt in (c) are the horizontal traces of the planes parallel to the rays 
which touch the surfaces along VR, VT, or RR, TT. See Theorem 4, 
Art. 327. 



XVII 



CAST SHADOWS 



417 





X (b 





yi/' 


\c: 


V 


\ 
r tif- 





,^s 


^\77 



X 



^cn\ 






' n'n 



tor 



fWf 



lnl \0o v 

1 1 j 




Examples. 1. A cone rests with its base on the ground; diameter 
of base 2", height 3 ", and the centre of the base 2j" from xy. 
Determine the shadow on the ground, the rays in plan and 
elevation making angles of 30" and 45" with xy. Indicate 
the line of separation, and the portion of the surface of the 
cone which is in shade. 

2. Suppose the cone in Ex. 1 to be cut by a horizontal plane 2" 

high ; find the shadow of the frustrum on the ground. 

3. Determine the shadow on the ground of the cone in Ex. 1 when 

resting with the axis vertical and the vertex on the ground. 
Show the line of separation. 

4. A cylinder 2" diameter, length of axis 3", rests with one end on 

the ground, the centre of the base being 2^" from xy ; if the 
rays are as in Ex. 1, determine the shadow on the ground. 
Indicate the line of separation. 

2 E 



4i3 PRACTICAL SOLID GEOMETRY chap. 



335. Problem. A given cylinder, axis AC, rests on 
the ground with its axis at right angles to the vertical 
plane. Determine the projections of its shadow on the 
planes of projection from given parallel rays. Show 
also the line of separation on the cylinder. 

We may regard the required shadow as consisting of 
three portions, one from the curved surface of the cylinder, 
and two others from the flat circular ends. If these three 
were found separately they would be seen to intersect each 
other ; only the outline is required for the shadow. 

To determine the shadow cast by the curved surface, 
draw the two tangential rays in elevation, touching the circle 
at /' and r, from which // and rr may be obtained by pro- 
jection. Then TT and RR form the line of separation for 
the curved part of the cylinder. 

Obtain r Q r Q and t e t v the shadows of RR and TT on 
the horizontal and vertical planes (Prob. 329) ; in determin- 
ing these use may be made of the fact that r r and t t are 
each equal and parallel to rr or //. 

To obtain the shadow cast by the circular end nearer to 
the vertical plane, determine c v the vertical trace of a ray 
through C, and with c x as centre and radius equal to that 
of the cylinder describe a circular arc ; it will touch e () t 1 at 
t x and meet xy in d . This arc is that portion of the shadow 
of the circle which falls on the vertical plane. 

Through d draw d d' parallel to the elevation of a ray ; 
then the arc DR is the only part of the circular end which 
casts a shadow on the ground. To obtain this shadow it 
will be sufficient to consider one point on DR between D 
and R (not shown), then by finding the shadow of this point 
we shall be enabled to draw the elliptical arc r d . 

The shadow from the other circular end consists of an 
elliptical arc extending from r to t Q as shown, and passing 
through a. Points on this arc may be found by taking a 
few points on the semicircle TNR and determining their 
shadows on the ground. 

Observe that r (J r t) , t Q f Q are tangents to the elliptic arcs at 



XVII 



CAST SHADOWS 



419 




i> '0 > 



r a t ft 

and that e t 1 touches the circle at t x ; note also 
that at and nn Q are tangents to the ellipse at a and n . 

The line of separation consists of the generators J?A, 
TT, together with the semicircles RNT, RDT. This is 
the line whose shadow gives the outline. 

Examples. 1. A cylinder, 2" diameter, length 2", rests on the 
ground along a generator which is at right angles to the vertical 
plane. The nearer end of the cylinder is " from the vertical 
plane. The rays, in plan and elevation, make angles of 30' 
and 45" with and are directed towards xy. Determine the 
shadow on the planes of projection ; indicate the line of separa- 
tion, showing the two parts of it which cast the shadow on the 
horizontal and vertical planes respectively. 

2. A semi-cylinder, 2.5" diameter, 2" long, rests with a semi- 

circular end on the ground, its curved surface touching the 
vertical plane, and its rectangular face parallel to the latter. 
Draw its plan and elevation and determine the shadow cast on 
the planes of projection, the parallel rays making 45 with 
xy in both plan and elevation, and being directed towards a v. 

3. A hemisphere 3" diameter rests with its curved surface on the 

ground, its flat face being horizontal, and its centre 2" in front 
of the vertical plane. Obtain the shadow cast on the planes 
of projection, the rays being as in Ex. 2. 



4 20 PRACTICAL SOLID GEOMETRY chap. 

336. Problem. A square slab as shown in the figure 
has a square hole cut through its centre ; to determine 
its shadow on the ground, the parallel rays being given. 
Also to indicate the line of separation. 

The shadow cast by the outer edges of the slab is easily 
obtained and requires no description. 

For the hole, the shadows from the upper and lower 
squares, supposing that they existed alone, would be 
/ O t o and iVyyo respectively. These squares intersect 
at two points, denoted by u , v and t , w ; from which we 
infer that the two rays which pass through them respectively 
must intersect the sides of both the upper and lower squares 
of the hole. Draw the plans of the rays through 7' and t 
intersecting the plan of the hole in v, u and t, w. 

It will now be obvious that the outline of the shadow 
from the edges of the hole is cast by TL and L V on the 
upper square, and UR, RJV on the lower, and is composed 
of the iimermost lines of the squares /<// ?y 0) / //i /i o . On 
the contrary the outline of the shadow of the outer edges 
of the slab might be obtained by first determining the 
shadow from the upper and lower squares and selecting the 
outer-most lines for the outline of the shadow. 

It should be noticed that the shadow from VM would 
be cast on the vertical face MNRQ of the hole and would 
be a straight line from U to M ; this is not shown. 
Similarly, the shadow from the line TO would be a straight 
line WO on the face ONUS. 

The line of separation therefore consists of the edges 
EF, FB, BC, CB>, DH, HE, and also the lines LM, 
MU, UR, RJF, 1 TO, OL. The portions TM, TO cast 
shadows on the internal surface of the hole, and there- 
fore do not contribute to the outline of the shadow on the 
ground. 

The student should learn from this instructive example 
that the points in which two shadows intersect may assist 
in tracing the line of separation, which always passes round 
the object in a circuit without break. 



XVII 



CAST SHADOWS 



421 



a 



I' d' 0' m'b' n! 



_i IIIIIMII 



:IHllll!lllllll ! llllll||[|llllllillllli; 



p' H, s' <f f r 



9 



\ 




Example. Copy the above plan and elevation half as large again 
as shown, and determine the shadow on the ground when the 
rays make 45" with xy in both plan and elevation. 



422 PRACTICAL SOLID GEOMETRY chap. 

337. Problem. To determine the projections of the 
line of separation on the surface of a given solid of 
revolution, axis vertical, the rays being parallel to the 
vertical plane, and hence to obtain the shadow of the 
solid on the ground. 

The solid considered is the same as that in Prob. 281. 
Let d be the centre for the arc deb'. 

Draw any radius o ' d' intersecting db' in s' ; with centre 
/ draw the circle through d '. This is the elevation of a 
sphere inscribed in the given figure, and touching it in a 
circle whose elevation is e'd'. Through / draw a diameter 
perpendicular to the given elevation of a ray, and meeting 
e'd' in/'. Then P is a point on the required line of separa- 
tion. 

For the ray through P evidently touches the sphere at P ; 
hence it must also touch the surface of revolution at P. 
(See Art. 285.) P 'is thus a point on the required line of 
separation. 

The plan of P may be found by describing a circle with 
s as centre, and diameter equal to e'd' ; a projector from /> 
intersects this circle in />, p, which are the plans of two 
points on the line of separation. 

Three or four other spheres should be taken, and the 
above construction repeated, a curve being drawn through 
the points thus found. 

The points A 7 " will be on the line of separation. Also, 
if a radius of be drawn at right angles to the elevation of 
a ray, the highest point F oi the curve will be found, and 
this will be on the outline in elevation. 

The shadow on the ground is obtained by considering 
rays which pass through the points on the line of separa- 
tion. For example, the points P give the shadows />, p . 

One half of the surface is illuminated. 

Examples. 1. Determine the shadow cast on the horizontal 
plane by the surface of revolution in Ex. 1, Prob. 282, when 
its axis is vertical, one end being on the ground. The rays are 
parallel to the vertical plane, and inclined at 45 . 



XVII 



CAST SHADOWS 



423 



\ /H. 




e'L 


p'flm 


1. s ', 


1/- ' '/ L -t k 


/A .'" / 


/=- = 



o 1 ' 



C 



X 



y 




Draw a line ab parallel to xy, 1 J" above it and 2^" long. With 
a/' as radius, and a and b in turn as centres, describe arcs 
commencing at b and a and terminating in xy. This is the 
elevation of a surface of revolution ; determine the shadow cast 
on the ground, the rays being parallel to the vertical plane, and 
inclined at 40. 



424 PRACTICAL SOLID GEOMETRY chap. 

338. Problem. To determine the shadow cast by the 
hexagonal head of the given holt on the cylindrical shank, 
from given parallel rays. 

In this example the plan of the shadow surface is also 
an edge view of the surface, so that the plan of the shadow 
is in this edge view. 

Draw the tangents ra, tb parallel to the given plan of 
the ray ; these represent the extreme rays, and the plan of 
the shadow on the shank is the semicircle rpt. It is 
evident that the shadow is cast from the portion ACB of 
the lower hexagon. 

We may choose any point Q (i.e. q, g) in ACB; then 
to determine the corresponding shadow P draw gf> the plan 
of the ray QP, and project from p to p' on to the elevation 
of the ray drawn through q. 

We ought now to select the rays which give the im- 
portant points. There is one through the angular point C 
which gives the angular point S of the shadow. 

Another is drawn through n in plan, and this determines 
n, a point on the outline in elevation. 

Next, to find the highest points in the two elliptic arcs of 
the shadow. For the arc PJVS draw on perpendicular to ac 
and uv the plan of a ray ; obtain vit' ; then U is the 
highest point of the curve RNS, and the tangent at u' is 
horizontal. A similar construction would determine P, the 
highest point in ST, the tangent at e being horizontal. 

The two extreme rays AR and B T illustrate the following 
theorem relating to shadows which fall on curved surfaces : 

Theorem. A ray which is tangential to the shadow 
surface is also tangential to the shadow ; and the plan and 
elevation of the ray are respectively tangential to the plan and 
elevation of the shadow. 

Thus b't', a'r touch the curves at f and r. 

Examples on Problems 338 and 339. 

1. Copy the figure of Prob. 338 double size, the edge AC being 
placed perpendicular to the vertical plane. Then work the 
problem, the rays making 45 with xy in plan and elevation. 



XVII 



CAST SHADOWS 



425 




C f & 







/ 

1 
1 


\ 


\ 
\ 


i 
1 



















2. Draw Figs, (a) and (//) to the dimensions given above, and de- 

termine the shadows of the heads on the shanks, as in Proh. 
33S ; the rays making 45 with xy in plan and elevation. 

3. Draw the projections (a) and (/') when the plans are turned 

through 45. Then work Ex. 2. 

4. The shank of a rivet is a cylinder l" diameter, and the head is 

a cone, diameter of base 1-^", height 1". Determine the 
shadow of the rivet on the planes of projections, the axis of the 
rivet being vertical and 2#" from the vertical plane : the rays 
in plan and elevation making angles of 30 and 40 with xy. 
Show also the shadow of the head on the shank, and the line of 
separation on the solid. Scale, double size. 

5. Work examples 1 to 4, supposing in each case the solid shank 

to be replaced by a thin hollow shank with the front half cut 
away and removed. 



426 PRACTICAL SOLID GEOMETRY chap. 



339. Problem. To determine the shadow cast by the 
head of the given rivet on the shank, and by the rivet on 
the planes of projection, the direction of the parallel rays 
of light being given. Also to indicate the line of separa- 
tion on the rivet. 

First, determine the shadow cast by the cylindrical 
shank on the ground, supposing- the vertical plane to be 
transparent ; it consists of the two tangents kk Q , // and the 
semicircles w / , kgl. 

Next, obtain the shadow cast by the head on the ground ; 
it consists of the circle, centre c Q , and its tangents v r Q , v Q L. 
Draw t Q t and r Q r parallel to the plan of a ray ; then VT 
and VR form the line of separation on the curved surface 
of the cone. The shadow of the cone is therefore that 
cast by the generators VR, VT, and the arc RFHT. 

Determine v x r v the shadow cast by VR on the vertical 
plane ; join zy . Then v x e Q t Q is the shadow cast by VT on 
the two planes of projection. 

Draw d Q d parallel to the plan of a ray ; then the arc 
DR casts its shadow on the vertical plane, and this shadow 
may be found by taking one or two points in DR and 
obtaining the shadows cast by them ; the construction is 
shown for one such point N, which gives the shadow n v 

The outline of the shadow on the horizontal plane is 
kutfltfjtfvjgk, and on the vertical plane is d^n^r^v^e^d^. 

To obtain the shadow of the head on the shank, draw 
the tangents Ih and //" parallel to the plan of a ray; these 
are the limiting rays, and the arc FH is that which casts 
the required shadow. 

Take any point q on fh and draw qp, the plan of a ray 
through Q. This ray is intercepted by the surface of the 
cylinder at a point on the generator whose plan is p ; draw 
the elevation of this generator, and through q draw q'p' 
parallel to the elevation of a ray and meeting the elevation 
of the generator at /', then p' is on the elevation of the 
shadow. 

Repeat this construction for a few other points on fh, 



XVII 



CAST SHADOWS 



427 




not forgetting to determine the important points /", s', g 
and the highest point on the curve. (See the remarks 
towards the end of the last problem.) The elevation of the 
shadow on the shank is the curve s'p'g'i'. 

The line of separation consists of two detached portions, 
each closed, viz. LSGIKL and VTHRV. 



428 PRACTICAL SOLID GEOMETRY chap. 

340. Problem. Having given any two lines AB and 
CD, and the direction of the rays of light ; to find the 
two points F and E in which a ray intersects both lines ; 
that is, to find the shadow of the one line on the other. 

Method. Find the shadows of both lines on any con- 
venient surface. Project the ray through the point where 
the shadows intersect. This ray will cut the given lines in 
the required points. 

In the figure the shadows aj> w c d on the horizontal 
plane are found ; these intersect in the point marked e / ; 
a ray through this point is seen to cut the given lines in the 
points Eand E. Then the point E is the shadow of AB 
on CD, and F is the point in AB which casts the shadow. 

Note.- The student should make careful note of the principle 
underlying this problem, and of the corresponding construction. The 
lines AB and CD may be curved. AB typifies the line of separation, 
and CD a selected line on the shadow surface ; the construction shows 
how to find the point where the shadow cuts CD. See applica- 
tions in following problems. 

341, Problem. Having given a line AB, and a thin 
circular plate, centre C, parallel to the ground, to determine 
their shadows on the ground. Also to obtain the shadow 
of the line on the plate, and the portions of the line which 
cast the shadows on the plate and ground. 

Obtain aj> , the shadow of the line on the ground, on 
the supposition that the plate is removed; find also the 
circular shadow, centre c , cast by the plate on the ground. 
These shadows intersect each other in e , f , and g Q , // . 
Thus the outline of the required shadow on the ground 
consists of the circle, centre c , and the lines a / and /> /i . 

Through e and g draw the lines e Q ef and g g/i parallel 
to the plan of a ray ; then, as in Prob. 340, the rays which 
meet the ground in e and g intersect the circular plate in 
E and G, and the line AB in F and H. Hence EG is 
the required shadow on the plate, cast by EH; and AE 
and BH throw their shadows on the ground. 



XVII 



CAST SHADOWS 



429 





As' 



\ 



X 



fj 



y 




iv? 


' 


e' 


'! ,r . 

!:! 9, 


\ ~^- 





341 




43o PRACTICAL SOLID GEOMETRY chap. 



342. Problem. The plans of a line EF and of a square 
pyramid VABCD are given, the points v, e, f being 
indexed, and the base ABCD resting on the ground. 
Determine the plan of the shadow cast by the line EF on 
the ground and on the pyramid, rs being the plan of a 
ray, the inclination of the rays being a. 

Draw xy parallel to rs, and project the elevations rs, 
e'f\ and*/. Obtain e Q / , the shadow of FF on the ground, 
supposing the pyramid to be removed ; also determine r a, 
the shadow of VA on the ground. These two shadows 
intersect in ///, and the corresponding point M on EF casts 
the shadow P on VA. 

The shadow of F will obviously fall on the face VAD, 
and to determine its projection join v f and produce to 
meet ad in o ; join ov, meeting f f in q. Then q is 
evidently the plan of the shadow of F oxs. the pyramid. 

The complete shadow of EF is ? u flq in plan. 

Examples on Problems 340 to 343. 

1. The x, y, z co-ordinates of two points A and B are (i", i", i") 

and (3", 3", 2"), and those of C and D are (2", 1", 1") and (1 ", 
3", 2"). If "the plan and front elevation of a ray make angles 
of 30 and 45 with the axis of Y, determine the projections and 
co-ordinates of the points E and F in which a ray intersects 
both of the lines AB and CD. 

2. Determine the projections of the line parallel to the vertical 

plane of ZX, which intersects the lines AB and CD in Ex. 1, 
and is inclined at 6o to the ground plane. 

3. The two ends A and B of a line 3" long are respectively 2" and 

1" from each of the two planes of projection. A horizontal 
circular plate, if" diameter, has its centre 1" vertically beneath 
the mid-point of AB. Determine the portion of the line AB 
which casts a shadow on the plate, the plan and elevation of a 
ray making 45 and 6o respectively with xy. 

4. An equilateral triangular plate, ih" side, is in a horizontal 

position 2" above the ground, the nearest corner A being i-|" 
from the vertical plane, and one side making 70" with the 
plane. A square plate, 2" side, is also in a horizontal position 
with one side at right angles to the vertical plane, one end of 
this side being 1" therefrom, 1" to the right of A, and 1" 
below it. Determine the shadow of the two plates on the 



XVII 



CAST SHADOWS 



431 




m/~... 







\ ^ ^-- // 


"s. 


/ \ 


K 



ground, and the shadow of the one on the other, the rays 
making 30 and 45 with xy in plan and elevation. 

5. A square pyramid, side of base 2", height 2^ ", has its base on 

the ground, one side making 30 with xy, the nearest corner of 
the base being 1" from xy. The end E of a line EF is 3V 
to the left of the vertex, J" from the vertical plane, and 3" 
above the ground ; the end F is 2" to the left of the vertex, 3'' 
from the vertical plane, and 2" above the ground. Determine 
the plan and elevation of the shadow of EF on the ground and 
pyramid, the rays being parallel to the vertical plane and in- 
clined at 40 to the ground. 

6. Describe two circles with a ar.d b as centres and radii " and 1", 

ab being 2". The circle with centre a is the plan of a circular 
plate 3" above the ground, the other circle being the plan 
of a sphere whose centre B is 1^" above the ground. Deter- 
mine the portion of the edge of the plate whose shadow falls on 
the sphere, the rays being inclined at 50 , their plans making 
io 3 with ab. 



PRACTICAL SOLID GEOMETRY CHAI'. 



343. Problem. The projections of a straight line RT, 
and a sphere, centre S, are given ; to determine the pro- 
jections of the portions of the line which cast their 
shadows on the sphere and ground respectively, the given 
rays being parallel to the vertical plane. 

The rays which touch the sphere will generate a cylinder 
the elevation of which is obtained by drawing the tangents 
g'l and tid parallel to the elevation of a ray ; this cylinder 
touches the sphere in a circle whose elevation is the diameter 
g'h', and this circle is the line of separation en the sphere. 

Make Iu=g'ti ; draw um perpendicular to xy and join 
////. Regard Im as the edge elevation of a plane ; this 
plane will cut the cylinder in an ellipse whose major axis 
LM is parallel to the vertical plane, the minor axis being- 
equal to the diameter of the cylinder. Now since it has 
been arranged that the plan of the major axis is equal to 
the diameter of the cylinder, the plan of the ellipse will be 
a circle the centre s 2 of which is projected from s 2 , the 
middle point of /;//. 

The shadow of the sphere, that is of its line of separa- 
tion, on the plane LM is the ellipse referred to ; hence 
the plan of the shadow is the circle, centre s.,. 

Next, obtain the plan of the shadow of the line RT on 
the plane LM. That is, draw r'r 2 and /?./ parallel to the 
elevation of a ray ; draw the projectors r 2 r 2 and t 2 \ to 
meet rr 2 and tt 2 in r 2 and t 2 respectively, then r/ 2 is the 
plan of the shadow of R T on the plane LM. 

The two shadows intersect in e 2 and/,. Now since e 2 is 
on the circle, centre s. the ray through E 2 touches the 
sphere ; also since e., is on r 2 t 2 the ray through E. 2 intersects 
RT. Similar remarks apply to /,. Hence the rays which 
pass through E 2 and F 2 respectively touch the sphere and 
intersect RT. "The projections of these rays are e 2 e 3 e, 
e^e' and / 2 / 3 / f^f, F 3 and F 3 being the points of 
contact with the sphere. 

Therefore the portion FFcasts its shadow on the sphere, 
and FT, FR cast their shadows on the ground. 



XVII 



CAST SHADOWS 



433 




Example. Describe a circle, centre c, radius i", and draw a line 
add to touch the circle at d ; make ad = \" , and db = \". 
Attach indices of 30, 20, and 15 to a, b, and c. Find the 
part of the line AB whose shadow falls on the sphere C, the 
rays being inclined at 45 and making 6o with ba in plan. 
Determine also the shadow cast on the sphere. Unit o. 1". 

2 F 



434 PRACTICAL SOLID GEOMETRY chap. 

344. Problem. The projections of a truncated cone 
with a circular slab placed centrally upon it are given ; 
it is required to determine the projections of the shadow- 
cast on the cone. The parallel rays are given. 

The shadow is cast from the lower edge of the slab, but 
the exact portion of this edge which casts the shadow on 
the cone is not so readily found as was the case with EQH 
in Prob. 339. The method now adopted is of general 
application, and might have been used in Probs. 338 and 
339, had it been necessary to do so. 

Obtain v the elevation of the vertex of the cone suppos- 
ing the truncated portion to be extended ; and by Prob. 334 
determine the projections of the two generators VT, VR, 
which form the line of separation on the cone ; this is done by 
first obtaining the shadow of the cone on the ground. Now 
draw the shadow cast from the lower circular edge of the 
slab ; it is the circle with centre c . This circle intersects 
the shadow of the cone in the points marked e / and g h Q . 

Draw e /e and g Q hg, the plans of the rays which meet 
the ground in e Q and g respectively ; then, by the principles 
explained in Prob. 340, these rays intersect the generators 
VR, VT in jFand H, and the lower edge of the slab in E 
and 67 respectively. Consequently the arc EG is that which 
casts the required shadow on the surface of the cone. 

Draw ov, o'v the projections of any generator O V, and 
draw ov the plan of the shadow of O V; let p g be the 
point where oz> intersects the shadow of the arc EG. 

Draw pQpq parallel to the plan of a ray, then the ray 
through p intersects OV in P and the lower edge of the 
slab in Q ; and therefore P is a point on the outline of the 
required shadow. The elevation /' may be found by pro- 
jecting from/ on ti o'v at /', or, as this is not very well 
conditioned, by projecting p to p Q ', then drawing p 'p' the 
elevation of the ray through p '. 

This construction should be repeated for other genera- 
tors similar to OV. The generator A V will give n on the 
outline in elevation. The two extreme rays EE, GH touch 



XVII 



\ 



CAST SHADOWS 



V' 
1 \\ 




the surface of the cone, and their projections are tangential to 
the projections of the shadow at / and g. The projec- 
tions of N, F, and H should be obtained as important points. 



436 PRACTICAL SOLID GEOMETRY chap. 

The preceding construction may be slightly varied in the 
following manner. Choose q, draw qp^ v Q p o, ov, the latter 
meeting qp in p. Draw p Q p ' and p Q 'p' to meet a projector 
from / in p' ; in this manner, however, JV would not be 
determined unless it were found accidentally. 

The elevation of the outline of the shadow on the cone 
is the curve f'n 'p'h', and the plan is the curve fnph, while 
HT and FR are portions of the line of separation on the 
conical- surface. 

Instead of taking generators such as O V, circles may 
be selected on the surface of the cone, in which case their 
shadows must be drawn and the points obtained in which 
these intersect the shadow of the lower edge of the slab. 
By this method, however, the limiting rays EF and GH 
could not be accurately determined. 

345. Problem. To determine the shadow cast by a 
given point P on a given oblique plane VTH, from given 
parallel rays RS. 

We must find the point where the ray through P meets 
the given plane. This may be done as in Prob. 199, or by 
either of the following methods. 

(a) Through p' draw p'tm parallel to r's', and draw mn 
perpendicular to xy. Then I' mn are the traces of an in- 
clined plane which contains the ray through P. 

Obtain nl, the plan of the intersection of the planes 
VTH, LMN ; and draw pq parallel to rs to meet In in q. 
Project from q to q . 

Then Q is the intersection of the ray PQ and the plane 
VTH, and is the required shadow on the plane. 

(b) Draw ////;/ parallel to rs, and nil' perpendicular to 
xy. Project from n to n ; join I'ri . Draw p'q parallel to 
r's ; and project from q to q. 

Then Q is the required shadow of P on the plane 
VTH. 

In this case a vertical plane containing the ray PQ 
has been substituted for the inclined plane of (a). 



XVII 



CAST SHADOWS 



437 




Note. The problem of finding the intersection ofa given lineand plane 
is generally best solved by applying the construction of either (a) or (b). 

Examples on Problems 344 and 345. 

1. A cone rests with its base on the ground ; diameter of base 

2\", height 5". It is cut by a horizontal plane which bisects 
the axis, and the upper portion is removed. ' A circular slab, 3" 
diameter and f" thick, rests centrally on the frustum of the 
cone. Determine the plan and elevation of the shadow cast 
from the slab on to the conical surface, the rays in plan and 
elevation making 30" and 50 with xy. 

2. The traces vt, th of an oblique plane make angles of 6o and 

40' with xy. A point P, in a plane through / at right angles 
to xy, is 2" from each plane of projection. Determine the 
shadow of P on the oblique plane, if the rays in plan and 
elevation make angles of 30^ and 45 with xy. 

3. Taking the plane VTH and the point /'as in Ex. 2, let PQR 

be an equilateral triangular plate, iV edge, in a horizontal 
position with PQ parallel to xy and directed towards the 
plane. Determine the shadow of the plate cast on the ground 
and on the plane, the direction of the rays being unaltered. 

4. An equilateral triangular plate ABC, 2" edge, has one edge 

AB on the ground at right angles to the vertical plane, the 
nearer end of the edge being 1" from xy ; the plate makes 40" 
with the ground. A line EF has its end E 1 .V to the left of 
AB, 2" above it, and 2" from the vertical plane. The end F 
is -i" to the left of AB, 1" above it, and 3" from the vertical 
plane. If the rays in plan and elevation make angles of 30 
and 45" with xy, determine the shadow of the line EF on 
the ground and on the triangular plate. 



438 PRACTICAL SOLID GEOMETRY chap. 

346. Miscellaneous Examples. 

1. A tetrahedron of iV' edge has its three lowest corners o. i", i", 
and 1.3" respectively above the horizontal plane. Show the 
shadow thrown by it upon that plane, assuming the rays of 
light to be inclined at 45 , and their plans to make an angle of 
45 with the horizontal trace of the plane containing the three 
given points. (1881) 

*2. Fig. (a). The plan of an octagonal prism, .V thick, with a 
square hole cut through it, is given. Assuming the height of the 
lower surface of the prism above the horizontal plane to be 1-i", 
obtain the complete outline of the shadow thrown on the hori- 
zontal plane. N.B. The rays of light are parallel, and their 
direction is given in plan and elevation. 0893) 

*3. Fig. (/;). The elevation and plan of a chimney are given ; ab is 
the plan of one of the parallel rays of light. Determine in plan 
the shadow cast upon the roof. Unit = 0.1". (1891) 

*4. Fig. (<). Draw the projections of the given cylinder and trun- 
cated cone four times the size of the diagram, and determine the 
outline of the combined shadow of the two surfaces, thrown by 
parallel rays, the direction of which is given, on the horizontal 
plane. The shadow of the cylinder on the cone is not required. 

(1890) 

*5. Fig. (</). The plan is given of a solid prism 1^" deep resting 
on the horizontal plane, r is the plan and r the elevation of 
a ray of light, to which the other rays are parallel. Show 
the shadow thrown on the horizontal and vertical planes. 

(1882) 

*6. Fig. (e) shows the plan and elevation of an inclined square plate 
out of which is cut a rectangular piece. Determine the por- 
tions of the shadow cast on the planes of projection. The plan 
and elevation of one of the parallel rays are shown. 
7. Determine the shadow cast on the horizontal plane from the cap- 
stan shown in Fig. (_/), p. 485. Take the rays parallel to the 
vertical plane and inclined at 35 to the ground ; also take the 
axis of the capstan 2" from the vertical plane. Show the line 
of separation on the surface. 

*8. Fig. (c). Copy the figure four times the size shown. Deter- 
mine the shadow cf the cylinder on the cone if the rays are 
parallel to the vertical plane and slope downwards to the right 
at 30 . Also obtain the shadow of the cone on the cylinder, 
if the rays are parallel to xy and directed to the left. 
9. A hollow cylinder 2" long, 3" and 2t>" external and internal 
diameters, is cut into two halves by an axial plane. One of the 
portions is placed with its end on the ground and the middle 
generator of its convex surface against the vertical plane. 



XVII 



CAST SHADOWS 



439 



Copy the figures doubl& siz& 




44Q PRACTICAL SOLID GEOMETRY chap. 

Determine the shadows on the concave surface, and on the 
planes of projection, the parallel rays making 45" with xy in 
plan and elevation. 

*10. Fig. (a). The figure represents a fluted column with a circular 
cap. Draw on the elevation the shadow thrown on the column 
by the cap. The direction of the parallel rays of light is 
given by arrows. ( 1 S97) 

""11. Fig. (p). The diagram, represents a horizontal square block, 
ABCD, intersecting a square-based pyramid VEFGH. (The 
overlapping portions of the solids are left dotted on the dia- 
gram.) Determine the intersections of the solids in plan and 
elevation ; and draw the outlines of the shadows thrown by 
them, one on the other, and by both on the horizontal plane. 
The arrows indicate the direction of the parallel rays. (1895) 

" ;t "12. Fig. (V). abc is the plan of the base of an octahedron lying in a 
horizontal plane iV above the ground ; fcjms that of a pyra- 
mid resting on the horizontal plane, its vertex (0) being 2.35" 
above the ground. The direction of parallel rays of light is 
shown in rjlan by r, their inclination to the horizontal plane 
being 44 . 

Show in plan the shadow of the octahedron cast on the 
ground and on the pyramid. Shade lightly all portions of the 
latter not illuminated. (1S96) 

*13. Fig. ((/). The given figure is the elevation of a square-headed 
bolt with a conical shank standing on the horizontal plane. 
The projections r, ;' of one of the parallel rays of light arc- 
given. Draw the bolt full size, and obtain the shadow cast on 
the horizontal plane and on the shank. (1887) 

*14. Fig. (e). Draw the plan, and determine the shadow cast on 
the ground by the trestle of which an end elevation is shown. 
The length of the top block is 4' 6", and it projects 6" beyond 
the legs at each end ; the legs are square in section. Let the 
plan and elevation of one of the parallel rays of light be parallel 
to each other and make 45 with xy. Scale -jV. 

r 15. Fig. (/"). A spherical segment rests on the top of a truncated 
hexagonal pyramid. The elevation and plan of the base are 
given. Draw the figure four times the given size, and determine 
the portion of the solid in shadow, as seen in elevation. The 
direction of the parallel rays of light is indicated. (1S88) 

16. A right circular cone, 3" high, rests with its base (3" diameter) 
on the ground. A sphere i-t" diameter has the vertex of the 
cone as its centre. Determine the shadow cast by the sphere 
on the cone, the rays being parallel to the vertical plane and in- 
clined at 45 to the ground, 



XV11 



CAST SHADOWS 



4M 



Copy the figures double size 



f 




<?' 



J-' / 



V s 



b'ci' \/!\\ c'd' 



I/aTTl 



I I \ \ 



c'd' 



e' f it g' . 




-Z'9- 



CHAPTER XVIII 

METRIC PROJECTION 

347. General explanation. We have seen how by an 
indexed plan the form or position of an object may be 
defined by one projection only. In this chapter we develop 
another method of representation by means of a single view. 
And in this case, as in the former one, the projection of 
the object is one that can be readily scaled for the purpose 
of ascertaining the dimensions of the parts. Each system 
is associated with a class of examples coming within its 
special province, and with which it is well adapted to deal. 

Thus the shape of an irregularly curved surface is well 
exhibited by the method of figured plans. AVe have a 
well-known example in maps with indexed contours, which 
indicate the configuration of the hills and valleys. A similar 
series of sections or contours is employed by naval architects 
in representing the shape of the surface of a vessel. 

Metric projection is well fitted to show the forms and 
dimensions of what may be termed j-ectatigular solids, such 
as are many examples of wood work ; they are bounded 
mainly by three systems of planes mutually perpendicular, 
intersecting in three systems of parallel lines in directions 
also mutually at right angles. The metric projection of 
such an object resembles a perspective view, and like the 
latter conveys a realistic impression of the form even to 
the uninitiated. The metric view may look distorted ; 
but there is compensation in that it is a scale drawing. 



chap, xvm METRIC PROJECTION 443 

348. Metric scales and axes. Let oa be the projection 
of a line OA on any plane. On OA set off any scale, say 
a scale of inches, and project the scale on oa. 

The latter scale may be used citlier (1) to ascertain the 
length of any line parallel to OA by measuring the length 
of its projection on the plane, or (2) to set off the length 
of the projection, knowing the length of the line. 

Such a scale is called a metric scale ; the plane of pro- 
jection is called the metric plane ; the direction oa in the 
plane is a metric direction ; any line in the plane and par- 
allel to oa is a metric line ; and any line of reference 
parallel to oa and in the plane is a metric axis. 

Suppose an object like a building brick is projected on 
the metric plane. The edges of the solid form three 
systems of parallel lines ; as each system will have its own 
scale, we shall require three scales in measuring all the lines 
of the projection. These are known as the trimetric scales. 

Let any three axes of reference OX, O Y, OZ be taken in 
space parallel to the three systems of lines, then their pro- 
jections ox, oy, oz on the metric plane are called metric 
axes of the projection. As was explained in Chapter VI L, 
the axes OX, OY, OZ serve to define the position of a point 
in space. Thus if the co-ordinates of a point P in space 
are X, Y, Z, the point may be reached from O by going 
first a distance X along OX, then a distance Y parallel to 
OY, and finally a distance Z parallel to OZ. And it is 
easily seen that the projection p of the point P may be 
plotted on the metric plane by stepping off first from a 
distance x along ox, then at the end of this a distance y 
parallel to oy, and finally a distance z parallel to oz, where 
x,y, z are the co-ordinates X, Y, Z, measured on the metric 
scales. 

Examples. 1. If a line 2.31" long he inclined at $S.f to the 
metric plane, find the length of its metric projection. 
Ans. 1. 81". 

2. Construct a half-size metric scale for a system of parallel lines 
inclined at 55" to the metric plane. The scale is to read 
inches and eighths of an inch. 



444 PRACTICAL SOLID GEOMETRY chap. 

349. Isometric projection the axes and scale. Sup- 
pose that in a trimetric system the three directions of the 
lines in space are all equally inclined to the metric plane, 
then the three scales become identical, and only one scale 
is required in measuring all the lines of the system in the 
metric projection. 

In this case we have an isometric or equal-scale system; 
the projection on the metric plane is called an isometric pro- 
jection, and the scale used is the isometric scale. 

The simplest example of such a solid is a cube; and the 
edges of a cube will evidently be all equally inclined when 
a diagonal of the solid, that is, a line joining two opposite 
corners, is perpendicular to the metric plane. This pro- 
jection of a cube may be readily determined by ordinary 
Descriptive Geometry, but is still more easily obtained by 
the method of isometric projection as follows. 

To project a cube isometrically. From considerations of 
symmetry we see that the three edges of the cube which 
radiate from the corner O farthest from the metric plane 
will project into three lines radiating from a point o at 
equal angles of 120 ; these are very conveniently drawn 
with the 30 and 90 angles of the set-square, as shown at 
ox, oy, oz in the figure. These lines define the three iso- 
metric directions ; and they, or any other concurrent parallel 
set, may be taken as isometric axes. 

Next, since the edges OA, OB, OC of the cube are equal 
and equally inclined, their projections are equal in length. 
Therefore mark off along ox, oy, oz three equal lengths 
oa, ob, oc. This length is at present undefined. 

Finally, to complete the isometric projection of the cube, 
we make use of the principle that parallel lines have parallel 
projections. Thus to complete the face of which oa, ob 
are sides, draw af and //parallel respectively to ob and oa, 
and intersecting in f. The corners d, e, g, with the sides 
radiating from them, are similarly determined. 

The outline of the projection is seen to be a regular 
hexagon. 



XVIII 



METRIC PROTECTION 



445 




W e must now find the length of the edge of the cube, 
of which the figure just obtained is the isometric projection. 
The corners A, B, C of the cube are evidently equi- 
distant from the plane of projection, for they are the ends of 
lines which all start from a common point O, and are equal 
and equally inclined. Thus the triangle ABC is parallel to 
the metric plane, and abc gives its true shape. Thus any 
side of the latter, say ab } is the true length of a diagonal of 
one face of the cube. 



44* PRACTICAL SOLID GEOMETRY chap. 

Draw aF, bF at 45 to ab ; then aF is a side of the 
square of which ab is a diagonal ; so aF is the required 
length of the edge of the cube. The following method of 
constructing an isometric scale is thus suggested : 

To construct an isometric scale. From any point a 1 draw 
a l F 1 parallel to aF, that is, with the 45 set-square; also 
draw rtj/j parallel to of, that is, with the 30 set-square. 
On a 1 F 1 set off a true length scale, and project the scale on 
a.f, by lines parallel to Ff, that is, lines drawn with the 
90 set-square. Then the projected scale a 1 / 1 is the re- 
quired isometric scale. Measuring oa on this scale, the edge 
of the cube is seen to be 2.95". 

To project a rectangular prism. Employing this scale, 
we have at the lower part of the figure, drawn half size, 
two isometric projections of an ordinary building brick 
9" x 4J" x 3". The process is quite simple. From any 
point draw three isometric lines ; set off along them from 
the point the length, breadth, and thickness of the solid, 
using the isometric scale ; complete the figure by drawing 
the system of parallel lines, as explained for the cube. 

To express the isometric scale numerically. Since the angles Fall, 
fall are 45 and 30, we have 

Fa sji fa 2 

I ~ ' and ^7 = "7= 
ah 1 an N /o 

af 2 \'2 sj~2 9 

hence -4^= -s == = .817 = nearly. 

af x / 3 1 v / 3 11 

This gives the ratio of the isometric length to the true length. 

We are not compelled to employ an isometric scale in 
setting out an isometric projection ; an ordinary scale may 
be used. But in this case the projection will be that of an 
object larger in the inverse ratio of the isometric scale. 
However, if this be recognised, and the ordinary scale used 
in measuring the projection, no error or ambiguity will 
result, and labour will be saved by using an ordinary scale. 



win METRIC PROJECTION 447 

350. Examples. Represent in isometric projection the following 

ten objects : 

1. A cube of 3" edge. 

2. A building brick 9" x 4!" x 3". Scale .1. 

3. A 24" square slab, 3" thick. Scale J. 

4. A 1" square prism, 3" long. 

5. A box without lid, 6" long, 4" wide, 2" deep inside, and -J" thick 
throughout. Scale -\. 

6. An instrument box with the lid open at right angles. Dimen- 
sions of box outside : length 6", breadth 4V, depth of box ij", 
depth of lid J". Thickness of wood at sides and ends ^"; at 
top and bottom \". Scale -Jr. 

7. The slab of Ex. 3, when pierced with 6" square hole through 
its centre. 

8. The chimney shown in Fig. (b), page 439. 

9. The trestle, Fig. (<), page 441. Scale -J. 

10. The column with cap, Fig. (a), page 441. 

11. Represent in isometric projection the three co-ordinate planes of 
Fig. 160, page 179, and in this view plot 

(a) The point A whose co-ordinates are (2", ii", 1"). 

(b) The point B whose coordinates are (2. 1.5", 1"). 

(c) The line CD joining the points (1", 2", 3"), (3", 2.5", 2"). 
(if) The line is/'' joining the ooints (1", - 2", 3"), ( 3", 

2-5", -2"). 

(e) The irregular tetrahedron whose angular points G, H, K, L 
are (2", iV, 1"), (1", 2", 3"), (1", 2.5", 1.5"), (1.5", 

5", 3- 5")- 
(/) The traces of the plane whose intercepts OA, OB, OC 
are 2", 3", and 4" respectively. 

12. A room is 24 feet long in the direction north and south, 18 
feet wide east and west, and 12 feet high. The following are 
points within the room : 

(a) A, situated 4 feet above the floor, 7 feet from the north 

Avail, and 5 feet from the west wall. 
(/') B, 3 feet high, 6 feet and 8 feet from the south and 

east walls. 
(<-) C, 4 feet below the ceiling, and 1 5 and 9 feet from 

the south and west walls. 
(d) />, 5 feet below the ceiling, and 3 and 12 feet from 

the north and east walls. 
Draw the room in isometric projection, and in this view 
plot the projections a, b, c, if of the points A, B, C, D. 

Nole. An isometric scale should be constructed and used in some 
of the above examples. Afterwards an ordinary scale may be 
employed to set off isometric dimensions. 



448 PRACTICAL SOLID GEOMETRY chap. 

351. Problem. Required the isometric projection of 
a cube of given edge which has a sphere of given radius 
in contact with the centre of one face, and a circle in- 
scribed in the same face ; a cone of given axis with its 
base inscribed in a second face ; and a cylinder of given 
length with its base inscribed in a third face. 

The cube. Let OA be the given edge. Determine On 
the isometric length of the edge. Set out the projection 
a, a, ... of the cube, as explained in the last article. 

For convenience refer to the visible faces as the upper, and the 
right and left front vertical faces. "We shall require the centres c ; the 
isometric bisectors dd ; and the diagonals aa of these faces. 

The sphere. Let the sphere rest on the centre of the 
top face, and let OB be its given radius. Obtain the 
isometric radius Ob, and set this up vertically from c to b. 
With b as centre, radius OB, describe a circle. This circle 
is the required isometric projection of the sphere. 

Note I. The diameters of the sphere and its projection are equal. 

The circle. Draw a circle DD of the given diameter 
OA, and circumscribe this by a square AA ; draw the 
diagonals AA, AA intersecting the circle in Jlf, M, N, N. 
Draw the equal perpendiculars MK, NK and obtain their 
isometric length Ok. Mark off this length at ak, ak, . . . 
from the corners of the upper face along the sides as 
shown, and through the points k, k, . . . draw the iso- 
metric lines intersecting in ///, ///, >i, n. The ellipse drawn 
through the eight points ;//, . . n, . . d . . is the required 
projection of the circle inscribed in the face. The lines 
dd may be called the isometric diameters of the circle. 

Note 2. Observe that this is an application of co-ordinates. The 
eight points M . . N . . D . . in the circle are referred to the sides 
of a circumscribing square A A. The co-ordinates are reduced by the 
isometric scale, and then plotted on the projection a, a of this square. 
Any plane curve can be similarly referred and plotted. And any ir- 
regular solid figure can be referred to a circumscribing rectangular prism, 
and its points then plotted on the isometric projection of the prism. 

The cone. Let O V be the given length of axis of the 
cone, and let the base be inscribed in the left front vertical 
face. Draw the ellipse inscribed in the projection of the 



XVIII 



METRIC PROJECTION 



449 




K A 



2 G 



45o PRACTICAL SOLID GEOMETRY chap. 

face in the manner just explained. Draw cv equal to Ov, 
and in the proper isometric direction, to represent the axis 
CV oi the cone. Draw the tangents to the ellipse from 
v, the projection of the vertex. 

The cylinder. Let the cylinder have its base inscribed 
in the right front vertical face of the cube, and let OE be 
the given length of the cylinder. Inscribe the ellipse in 
the projection of this face ; this is the projection of one 
end of the cylinder. To obtain that for the other end, 
conceive the ellipse to be moved parallel to itself in the 
isometric direction proper to the axis CE of the cylinder, 
through a distance Oe equal to isometric length of the axis. 
The construction suggested is to draw isometric lines 
through a number of points on the first ellipse, marking off 
on them lengths each equal to Oe ; we thus obtain points 
on the second ellipse. Some of these lines are shown at 
// in the figure. The projection of the cylinder is com- 
pleted by drawing the common tangents to the two ellipses. 

Note 3. Second /net hod of projecting a circle. 

Required the isometric projection of a circle, centre c', 
radius OE, the plane of which is parallel to the right front 
vertical face of the standard cube. 

All the lines dd, mm, nn on any face of the cube may be drawn 
with one or other of the 30 , 6o, and 90 edges of the set-square. 

First, through c draw the isometric lines dd, dd (whose directions 
are readily seen by reference to a standard cube) ; and along them set 
off the four isometric radii e'd, each equal to Oe. Next draw the lines 
through c in the directions of the major and minor axes. On the 
former (that bisecting the acute angles between dd, dd) set off cm, cm 
each equal to the true radius OE. Finally through m, m draw the 
isometric lines intersecting in 11, n. 

Then mm, nn, are the major and minor axes, and vinmn is an in- 
scribed square. We thus obtain the projection by drawing the iso- 
metric diameter, the major axis, and the inscribed square. Before 
sketching the ellipse through the eight points thus found, it is well to 
draw short lengths of the tangents at these points as shown. 

Note 4. If an ordinary scale be used to set off isometric dimensions, 
the radius of the circle which represents a sphere, or the semi-major 
axis of the ellipse which represents a circle, are larger than the actual 
radius in the inverse ratio of the isometric scale. 



xviii METRIC PROJECTION 451 

352, Examples. Represent by an isometric projection : 

1. A sphere of 1 \" diameter. 

2. A circle 2" diameter. 

3- A cone, diameter of base 2.2", length of axis 1.7 ". 

4- A cylinder 2.5" diameter, .6" long. 

5. A grindstone, 24" diameter, 5" thick, with a 6' square hole 

through its centre. Scale *. 
6- A rectangular slab of stone, 18" long, 12" broad, 3" thick, with 



a circular hole 5" diameter through its centre. Scale 



7. The frustum of a cone, the diameters of the ends 2.5" and 1. 5", 
length 1 ". 

8. A square pyramid, base 2^" side, axis 3V long. 

9. A hexagonal pyramid, base 1.5" side, axis 2.5" long. 

10. The frustum of the pyramid, Ex. 9, obtained by a plane bisect- 
ing its axis at right angles. 

11. A hemisphere, 3. 1" diameter. 

12. A hemispherical bowl, 20" diameter inside, J" thick. Scale |. 

13- A semicircle, and a quarter circle, 3" diameter. 

14- A half cone, 2.5" base, 2" axis, obtained by a cutting plane 
through the axis of the cone. 

15. A half cylinder, 2" base, 1.5" long, obtained by a plane con- 
taining the axis. 

16. A quarter cone and a quarter cylinder, of dimensions as in 
Exs. 14 and 15, obtained by planes containing the axis. 

17- A quarter sphere, and an eighth of a sphere, obtained by two 
or three mutually perpendicular planes through the centre of a 
sphere 4" diameter. 

18. A cone, base i\" diameter, axis 1^-", resting with its base con- 
centrically on one end of a cylinder, i^" diameter, 2" long. 

19- A ring, 2" internal diameter, of section -J" square. 

20- A ring, 2" internal diameter, of circular section, \" diameter. 

21- A regular tetrahedron of 3" edge. 

22. A regular octahedron of 2" edge. 

Hint. In examples like 21 and 22 there is a choice as to 
which three perpendicular lines connected with the solid shall be 
taken for the three principal directions. In the tetrahedron we 
might select the base and altitude of one triangular face, and the 
direction perpendicular to the face. In the octahedron the three 
mutually perpendicular diagonals of the solid might be chosen. 

23. A cone of indefinite length having a vertical angle of 50 . 

24. The surface of revolution, page 337, double the size shown. 

Hint. Employ inscribed spheres in Exs. 23 and 24. 

Note. In the above examples the student may use either an 
isometric or an ordinary scale in setting off dimensions in the 
isometric views. 



452 PRACTICAL SOLID GEOMETRY chap. 

353. Problem. To determine the isometric projection 
of the solid given in plan and elevation. 

Draw the isometric axes a x x, a x y, a x z, and set off a x b v 
a x d v and a x e x respectively equal to the isometric lengths (as 
obtained by a scale) of the lines ab, ad, and a'e . In like 
manner obtain the projections of the other edges parallel to 
AE. The four arcs joining the upper extremities of these 
projections must now be determined. 

Divide the circular arc e'f into a number of equal parts, 
say four. Join e x f x . Consider one of the points, say r ; 
draw r'ri perpendicular to e'f . Set off e x n and nr x re- 
spectively equal to the isometric lengths of e'ri and n'r ; 
this determines r v and the corresponding point on the arc 
parallel to e x r x f x is obtained by drawing r x r x equal and 
parallel to a~d v By a repetition of this construction the 
four arcs may be obtained. 

To determine the isometric projection of the cylindrical 
portion of the solid we may proceed as directed in the 
last problem or as follows : Suppose that the cylinder is 
circumscribed by a square prism which extends to the 
base of the solid ; its plan and elevation are shown in 
the figure, and its isometric projection may be thus ob- 
tained 

Draw the diagonal b x d v and from its middle point i\ set 
off i x t x and i x u x respectively equal to it and in. ( TU is 
not shortened by projection.) 

Make i x w x and i x v x respectively equal to the isometric 
lengths of i'w and i'v. It will then be seen that the three 
parallelograms with q, w v and v x for their centres can be 
completed. 

The ellipses which are the projections of the circular 
ends of the cylindrical portion of the solid must now be 
inscribed in the parallelograms of which w x and v x are the 
centres. Make w x g equal to the true length of the radius 
of the cylinder, and complete both ellipses in the manner 
explained in Prob. 352 ; two common tangents to the 
ellipses will complete the required projection of the solid. 



XVIII 



METRIC PROTECTION 



45 J 



C 



a' 



d 




I 




IV 



a 



r 



b 



Examples. Represent in isometric projection 

1. A sphere of 3" diameter penetrated by a cone of base 3" diameter, 

axis 4" long, the centre of the sphere being at the middle point 
of the axis of the cone. 

2. The wedge-shaped figure formed from a cylinder 3" diameter, 

3" long, by two sloping planes containing a diameter of one 
end, and touching the circle at the other end at opposite points, 
thus cutting away the sides. 
The instrument box of Ex. 6, p. 447, with the lid opened at 



3. 



130 



Scale 



\ or ' 



4. One of the four quarters of a hollow sphere, formed by two 

perpendicular planes containing a diameter. External and 
internal diameters 3" and 2". 

5. The separated parts (V) and (</) of the dovetailed joint, p. 457, 

represented (fitted together) by ordinary projection at (a) and 

6. The bolt with hexagonal head, p. 425, to double the size shown. 

The upper surface of the head is spherical. 

7. The bolts with round and square heads, p. 425, to the dimen- 

sions given. The upper surface of the square head is spherical. 
Note. In working these examples, the use of an isometric scale is 
optional. 



454 PRACTICAL SOLID GEOMETRY CHAP. 

354. Problem. Having given a set of trimetric 
axes, to determine the corresponding trimetric scales. 

Let ox, oy, oz to the left be the given axes. Take any 
point d in oy, and draw df, fe, ed respectively perpendicu- 
lar to ox, oy, oz. Now ox, oy, oz are the projections of 
three lines OX, O Y, OZ in space mutually perpendicular, 
and df, fe, ed are the traces of the planes YZ, ZX, XY on 
the plane of projection, or on a parallel plane. 

On ed as diameter, describe a semicircle cutting oz in o Q ; 
join o e, o Q d. Thus eo Q d is a right angle, and the triangle 
eo d is the rabatment of EOD into the horizontal position 
about ed. Thus eo Q and do are the true lengths of the 
lines of which eo, do are the metric projections. 

Again, the right-angled triangle of which the line fgo is 
the plan, is shown rabatted at fgo l ; so that fo x is the true 
length of the line of which fo is the metric length. 

The construction for the trimetric scales is thus at once 
obtained. Draw Ox, Oy, Oz parallel to the given axes, and 
draw OX, OY, OZ respectively parallel to o Q e, o () d, o x f. On 
OX, OY, OZ set off true length scales, and project these 
scales on Ox, Oy, Oz by lines parallel to oo rj oo , oo v 

The scales thus projected on Ox, Oy, Oz are the 
trimetric scales required. 

Note. Having given the trimetric scales 1, in, n, or their ratios 
I : in : n, to determine the trimetric axes. 

The values /, m, n are the ratios of the projected to the true lengths. 
Construct a triangle xyz, the sides yz, zx, xy of which are respect- 
ively equal or proportional to / 2 , nfi, and n 2 . Determine o the 
centre of the circle inscribed in this triangle, and join ox, oy, and oz ; 
these three lines are the required trimetric axes. 
It can be shown that P + in- + n- = 2. 

Examples. 1. The angles xoz xa&yoz between given trimetric axes 
are respectively 135 and 120. Construct the trimetric scales. 
Measure the representative fractions of these scales. 

.4ns. I, in, n for the scales ox, oy, oz are .S38, .932, .649. 
Note. The trimetric axes in this example may be drawn with the 

90 , 45 , and 30 edges of the set-squares. 
2. If the trimetric scales /, in, n are in the ratios I : J : |, determine 
the trimetric axes and measure the angles between them. 
Ans. lof, 138.5, 114.5- 



XVIII 



METRIC PROJECTION 



45! 




Examples on Problems 354 to 357. 

Represent in trimetric projection the following six objects, the metric 
axes being such that oz is vertical on the drawing-paper, the 
angle xoz 135 , and the angle yoz 120'. All the lines maybe 
drawn with the ordinary set-squares. 

1. A cube of 3" edge, with a circle inscribed in each visible face. 

2. A building brick 9" x 4^" x 3". Scale \, 

3. A sphere of 2" diameter. 

4. Three lines, each 3" long, which mutually bisect one another at 

right angles. 

5. The rivet on p. 427 to double the size shown. 

6. The instrument box of Ex. 6, p. 447, with the lid open at 120 . 

7. Refer to Art. 357 and the figure on p. 459. Draw the axes oa, 

ob, oc respectively horizontal, vertical, and with the 30 edge of 
the set-square. Take the scales for lines parallel to oa and ob 
full size, and the scale for lines parallel to oc half size. Then 
draw the projection of a cube of 2.5" edge, with a circle in- 
scribed in each visible face. 

8. Take the axes and the scales as in Ex. 7, or alter them in any 

manner that may seem desirable, and represent in metric pro- 
jection the model or the three planes of reference as shown in 
Eig. 161 (/>), making OX, O V, and OZ each 4". 

In this view plot the point A, the line CD, and the plane 
ABC of Ex. 11, p. 447. 

9. Take the axes and scales as in Ex. 8, and draw the metric pro- 

jection of the steps, Eig. (a), p. 461. 



456 PRACTICAL SOLID GEOMETRY chap. 

355. Problem. Having given the trimetric axes ox, 
oy, oz, to draw the trimetric projection of a cube of given 
edge with a circle inscribed in one face. 

First determine the trimetric scales as in Prob. 354. The 
axes and scales in Prob. 354 are used again. 

To project the cube, set off the lengths oa, ob, oc for the 
edges, each to its proper scale. Complete the figure by 
drawing the parallel lines as in isometric projection. 

Let the circle be inscribed in the right front vertical 
face ; its projection is the principal ellipse inscribed in the 
projection of this face. The ellipse may be set out as in 
Prob. 88, or as follows : 

Draw the two diagonals ac, of, and through their intersec- 
tion / draw the two trimetric diameters gs, pr parallel to oy, 
oz ; then /, g, r, s are four points in the ellipse. Draw the 
separate figure, centre I, and then locate n by plotting the 
co-ordinates om, mn equal to OM, MN reduced by the 
scales. The other points ti, v, w on the diagonals are then 
determined by drawing the projection of the inscribed 
square. The ellipse can now be traced through the eight 
points found. The tangents at all these points are known. 

Note. Observe that the diameters itv, nw which lie on the dia- 
gonals ac and of are not the major and minor axes of the ellipse, as was 
the case with isometric projection. The major axis is the line through 
7 perpendicular to ox, of length equal to the true diameter of the circle, 
for this line is parallel to the plane of projection. And generally, a 
line in or parallel to one face, say xoy, is parallel to the metric plane, 
when its projection is at right angles to the perpendicular axis oz. 

356. Problem. Two views (a) and (b) of a dovetailed 
joint are given. To draw a trimetric projection of each 
portion of the joint, the trimetric axes and scales being 
taken the same as in Prob. 354. 

The required projections are shown at (c) and (d). 

After the detailed descriptions previously given, the method 
of obtaining these should not require further explanation. 
The lines EH and FL, not being parallel to a trimetric 
axis, their projections cannot be scaled. Their ends are 
located by the method of co-ordinates. See the line KH. 



XVIII 



METRIC PROJECTION 



457 




M 



x kjf 


X" 


\ 

V 

1 *' 




\ / 


>< 



c 



s' 


t' 


0' 


H 


9' 



" 



f e r o 



(a) 



I h 

(b) 

356 




458 PRACTICAL SOLID GEOMETRY chap. 

357. Generalisation of the foregoing methods. The 
object of metric projection is to define a form having three 
dimensions by one pictorial view drawn to scale. In the 
examples hitherto considered this has been effected by 
orthogonal projection. We may, however, remove the restric- 
tion that the projection shall be orthographic, and still 
secure the object in view, and at the same time gain con- 
siderably in freedom. 

Radial projection is not permissible, because a perspective 
view cannot be scaled ; but we may employ oblique parallel 
projection, since in this case parallel lines project into parallel 
lines, all to the same scale. Now if the projectors, instead 
of being perpendicular, may be inclined, and in any 
direction, we gain two degrees of freedom in arranging for 
the projection. We may now select what two angles we 
like for the latitude and longitude, or the altitude and 
azimuth, so to speak, of our projectors. 

Let us now see how this helps us. In trimetric ortho- 
gonal projection we may choose the axes, and then we 
require to determine the corresponding scales. In trimetric 
oblique projection, with two more degrees of freedom, we 
may choose the axes and any two of the scales ; or what is 
equivalent, the axes and the ratios /:///: ;/ of the scales, 
leaving the absolute scale to be determined by a geometrical 
construction, if this should ever be required. But in 
practical applications we are not concerned with the 
absolute ratio which the size of the object bears to the size 
of its projection ; thus in isometric projection we generally 
use the full size instead of the isometric scale, and do not 
trouble ourselves as to how much bigger the object would 
have to be in order to actually project into the view drawn. 
Discarding consideration for the absolute scale, the general 
proposition may be thus stated : 

Proposition. /// trimetric oblique projection we may take 
the three axes in any directions we like, and 7ve max take 
the three scales whatever we like, and we shall have a true 
projection of the object, or of one similar in form. 



XVIII 



METRIC PROJECTION 



459 




As a simple illustration, draw oa, ob equal and perpen- 
dicular to one another, and draw oc in any direction and of 
any length. Complete the figure 
as shown by drawing the series of 
parallel lines. Then we are at 
liberty with perfect propriety to 
take this figure as the projection 
of a cube. Thus let the cube 
rest on the plane of projection, 
with one face coinciding with 
oadb. The line oc is the projec- 
tion of the perpendicular edge oC. 

Therefore Cc must be one projector, and the oblique pro- 
jectors are parallel to Cc. But generally it would not be 
so easy as in this case to locate the position of the object 
in space in regard to its projection. We may, however, 
employ the projection without solving this latter problem. 

We may remove another restriction. The three- 
directional system of lines in space need not be perpen- 
dicular to one another, but may have any directions. The 
proposition in its most general form may then be stated : 

Theorem. Let OA, OB, OC be any three lines of definite 
length in space, and oa, oh, oc any three lines of definite length, 
in one plane : then the former lines ran be projected into a 
figure similar to the latter by parallel projection. 

In applying this general method to practical cases, some 
regard must be had to the effects of distortion. Any pro- 
jection would appear right if viewed in a direction parallel 
to the oblique projectors. But a picture is generally looked 
at from somewhere near the front ; if the projectors were 
very oblique this front view would appear very distorted; so 
the axes and scales must be kept within reasonable limits. 

As in ordinary trimetric projection we can plot the pro- 
jections of irregularly situated points or lines from their 
co-ordinates, and circles may be plotted by projecting the 
circumscribing square or parallelogram, and drawing the 
principal inscribed adlipse. 



46o PRACTICAL SOLID GEOMETRY chap. 

358. Miscellaneous Examples. 

*1. Fig. (a). Draw the isometric projection of the object represented 
orthographically. A section on AB is shown. Scale T ^. 

(1896) 
*2. Fig. (b). Draw the letter P from the dimensions given in the 
sketch. Assuming the thickness of the material from which it 
is cut to be y", make an isometric view of the letter. 

N.B. An isometric scale is not to be used. (1S93) 

*3. Fig. (t). Make an isometric view of the cross. {^19) 

*4. Fig. (d). Make an isometric view of the bolt head, the elevation 
and half plan of which are given. 

N.B. An isometric scale is not to be used. Lengths to be 
transferred direct from the given figure to the isometric lines. 

(1886) 
*5. Fig. (e). The two lines ad, ac form one of the right angles of a 
face of a cube of 2|" edge. Complete the plan of this cube. 
Unit o. 1". (1 89 1) 

*6. Fig. (/). Two elevations of a gable cross are given. Make an 
isometric view of the cross. An isometric scale is not to be 
used. (1892) 

7. Make an isometric view of the bolt, Fig. {d), p. 441, drawing it 

full size. An isometric scale is not to be used. (1887) 

8. Particulars of a trestle are given in Ex. 13, p. 440. Represent 

the trestle in isometric projection. Scale \. Use of isometric 
scale is optional. 

9. Draw an isometric projection of the desk shown in figure, p. 

203. Use of scale optional. 

10. On each face of a cube of 2" edge stands an equal cube. Make 

an isometric view of the solid formed by these seven cubes. An 
isometric scale is not to be used. (1878) 

11. Draw the isometric projection of the object in Fig. (a), p. 441, 

standing on the horizontal plane, one isometric plane to be 
taken parallel to a diagonal such as fg, passing through two 
opposite edges of the fluted column. An isometric scale is not 
to be used. (1897) 

12. Draw three lines meeting at a point and making angles of 120, 

1 30, no Q , with each other. These lines are the plans of the 
edges of a cube of 3" edge. Complete the plan of the cube and 
draw its elevation on any plane parallel to no side or diagonal. 

(1879) 



X V 1 1 1 



METRIC PROJECTION 



461 



Copy die figures dvub/e sc^e 

y 



\ 



(a) 



/ is' 

-6'-A 

i 



1 1 




^r 


+ ] 








ix 




CHAPTER XIX 



MISCELLANEOUS PROBLEMS 



359. The five regular polyhedra. It is shown in 
treatises on pure solid geometry that there are five, and 
only five regular polyhedra. These are the tetrahedron, 
cube, octahedron, dodecahedron, and icosahedron ; see the 
Appendix, Definitions 18 to 22. We have already had 
occasion to represent the first three in projection ; the 
remaining two will now be considered. 

The regular dodecahedron, Fig. (1). This solid has 
twelve equal and regular pentagonal faces. The projection 
easiest to determine is that on the plane of one face. 

To obtain this, draw the regular pentagon aaaaa to re- 
present the bottom face. On two adjacent sides construct 
regular pentagons, and regard these as the rabatments of 
two adjacent faces. Thus aB , aB are the rabatments of 
the same edge about the two axes aa, aa. From B w B 
draw perpendiculars to the axes intersecting in b ; then 
ab is the plan of a sloping edge. The plan may now be 
completed from considerations of symmetry 

From the corners a, a, a, a draw the plans ab of the 
other four similarly-situated edges AB. For the top face, 
draw the regular pentagon ddddd, with sides parallel to aaaaa 
and circumscribing: the same circle. Draw the five lines 
dc each equal to ab, to represent the five edges sloping 
from the top face. Complete the plan of the solid by 
drawing the outline, which is a regular decagon. 



chap, xix MISCELLANEOUS PROBLEMS 



S B, 



4"3 




The distances of the points B, C, and D from the lower face A are 
readily found since we have the plans, and we know the true lengths of 
the lines AR, BC, and CD : we can thus draw an elevation on any 
vertical plane. 

If we draw an elevation on a plane parallel to a diagonal AD of the 
solid (a diagonal being a line which joins opposite corners), then we can 
measure the length of the diagonal ; and from the elevation we may 
project a plan with the diagonal vertical. This plan might also be 
drawn first-hand from considerations of symmetry. 

The regular icosahedro/i, Fig. (2). This solid is bounded 
by twenty equilateral triangular faces. Each angular point 
of the solid is the common vertex of five triangles, the 
bases of the triangles forming regular pentagons. We shall 
obtain the plan of the solid when a diagonal is vertical. 

1 )raw two regular pentagons b, b . . . c, c . . . circum- 
scribing the same circle and with sides parallel. Join all 
the points b and c to the centre, as shown in the figure, and 
also join the adjacent points b and c so as to obtain the 
regular decagonal outline. This figure is the projection of 
the solid on a plane perpendicular to the diagonal AD. 

We may obtain the heights of the various points, the length of the 
diagonal, the distance between parallel faces, and the projection on the 
plane of a face, by methods suggested for the dodecahedron. 



464 PRACTICAL SOLID GEOMETRY chap. 

360. Problem. To determine the spheres inscribed in 
and circumscribing a given regular polyhedron. 

Let the given solid be a regular tetrahedron, of which 
abed in the plan. 

The inscribed sphere. General method. First draw a 
view of the solid on a plane perpendicular to an edge, so as 
to project two adjacent faces in profile. Next obtain the 
projection d of the centre of the solid. Finally with centre 
d draw the circle which touches the profile projections of 
the two faces. This will be the projection of the inscribed 
sphere. 

Thus take xy perpendicular to ab, and draw the elevation 
a'b'c'd'. Draw the perpendicular d ' m and the bisector an 
to intersect in d ' . With centres d and o, radius dm', draw 
two circles. 

These circles are the projections of the inscribed sphere. 

The circumscribing sphere. General method. First 
obtain the projection of the solid on a plane equidistant 
from the centre O and two corners. Then with centre d 
draw the circle through the projections of the corners. 
This will be projection of the circumscribing sphere. 

Take ay parallel to ocd, and draw the elevation a'b'c'd. 
Obtain d as before. With centres d and o, radius o'd\ draw 
two circles. 

These circles are the projections of the circumscribing 
sphere. 

Note. To determine the spheres inscribed in, and circum- 
scribing a given irregular tetrahedron. 

For the inscribed sphere, first determine the three planes which bisect 
any three dihedral angles of the solid. Then determine the sphere 
which has its centre at the point of intersection of the planes, and 
which touches any face of the solid. 

For the circumscribing sphere first obtain the three planes which 
bkect at right angles any three edges of the solid. Then determine the 
sphere which has its centre at the point of intersection of the planes, 
and passes through any coiner of the solid. 

It is seen that this problem is the same as that of finding a sphere 
which shall touch four given planes, or contain four given points. 



XIX 



MISCELLANEOUS PROBLEMS 



465 




361. Problem. To determine any regular polyhedron 
inscribed in or circumscribing a given sphere. 

Let the circle, centre o v be a projection of the given 
sphere, and let the required figure be a tetrahedron. 

First work the last problem for a tetrahedron of any 
assumed edge, that is, draw Fig. 360. 

Through o x draw lines parallel to d'm' and a'n . 

The inscribed tetrahedron. Through d x draw d x a v d x c x 
parallel to d'd, d'/, and join c x a v which will be parallel to 
c a . 

The circumscribing tetrahedron.- Draw the tangents a.^d., 
a.,c, parallel to ad', dc, and through d., draw d c.> parallel 10 
d'c'. 

The lengths of the edges of the inscribed and circum- 
scribing tetrahedra are d x c x and d./., respectively. 

A similar method is used for any regular solid. 

2 H 



466 PRACTICAL SOLID GEOMETRY chap. 

362. Trihedral angles and spherical triangles. Let 

O be the apex of any trihedral angle, and suppose a section 

of the angle to be made by any 

-B/ spherical surface whose centre 

s^\\ i s at O- 1 ne sections of the 

s' \ three faces will be arcs of ereat 

s' \ /y circles, forming a spherical tri- 

/^l_ J 4 angle ABC on the surface of 

^^^^ / / the sphere. 

^"~~\/ We may let A", B, C stand 

^4""^ for the angles, and a, b } c for 

the sides. By the former we 
mean the angles between the tangents to the curved sides 
at the corners, and by the latter the angles subtended by the 
curved sides at the centre of the sphere. Thus the angles 
A, B, C\ and the sides a , b, c" of the spherical triangle 
are respectively equal to the three dihedral angles and 
the three plane angles or faces of the solid trihedral angle. 

The polar triangle. If through the centre of the sphere, lines OA v 
OB v OC x be let fall respectively perpendicular to the faces OBC, OCA, 
OAB, and all directed outwards (or all inwards), meeting the spherical 
surface in A v B v C v we have a second trihedral angle formed, with its 
corresponding spherical triangle A,B^C,. The latter is called the 
polar triangle of the triangle ABC. 

We have not space to establish the well-known relations between 
the sides and angles of the triangles ABC, A l B l C v therefore we shall 
merely state them. Denoting the sides and angles of the polar triangle 
A l B l C l by tfj , /; t , c x ; A, B^, Cj, it may be shown that the angles 
of the one triangle are supplementary to the sides of the other ; that is, 
we have 

<4 1 B =l&o a -a a l =i8o -A 

B^ = i8o" - i> b 1 =i8o-B 

C l =iSo-e * 1 =i8o -C 

The relations between the two triangles are reciprocal; each triangle 
is the polar of the other. 

The principal use of the polar triangle is in the solution of spherical 
triangles. It reduces, to one-half, the number of cases necessary to be 
dealt with. 

Problems on spherical triangles are equivalent to problems on tri- 
hedral angles. In solving problems on the former we shall require to 
draw projections of the latter. 



XIX 



MISCELLANEOUS PROBLEMS 



467 




363. Problem. To solve a spherical triangle or tri- 
hedral angle, having given : I. Three sides, or three angles. 
II. Two sides and the included angle, or two angles and the 
adjacent side. III. Two sides and one angle opposite to 
one of them, or two angles and a side opposite to one. 

Case I. Let the three given sides be a", //', c. 

Begin by drawing a development of the trihedral angle. 
That is, draw any circle, centre O, and set off the angles 
B x OC, CO A, AOB x equal respectively to the given sides 
a', />', c. Through B v B x draw perpendiculars to OC, OA 
intersecting in /; ; then b is the plan of the point B when 
the two outer sides or faces a and c of the solid angle are 
turned into their true positions, about <9Cand OA. 

Join Ob. We have now obtained a projection of the 
solid angle on the face OAC. The projection AbC of 
the spherical triangle is also shown. Several simple methods 
of setting out the elliptic arcs Ab, ^Cwill suggest themselves 
to the student. 

Through B v B x draw tangents to the circle, that is 



468 PRACTICAL SOLID GEOMETRY chaf. 



perpendiculars to OB v 0B V meeting OC and OA pro- 
duced in //and K. 

The three required angles can now be found. The 
angle B is obtained by the rabatment HB ? Kol the triangle 
HBK. The angles A and C are equal" to bNB^ bMB 2 , 
determined by the rabatments of the triangles bNB, bMB. 
The problem is thus solved. 

If the three angles A, B, C were given, the problem 
could be reduced to the one just worked, by means of the 
polar triangle. We should first subtract each angle from 
i8o, and thus obtain the sides of the polar triangle. 
Then by the construction above we could find the angles 
of the polar triangle, the supplements of which would give 
us the required sides of the original triangle. 

Note I. Observe the properties of the figure. 

There are five rabatments of the point B about five different axes, 
from which we have the relations 
MB l ^MB 2 ; NB x = NB t \ &B 2 = bB 2 ; HB X =HB Z ; KB X =KB % . 

Since the plane HBK'xs, perpendicular to the line OB, the trace HK 
is perpendicular to the projection Ob. Hence also B :i falls on Ob, 

Note 2. A simple and effective model can be made by cutting out 
the shape OB^HKB^O in paper, drawing on it the arc B\CAB V and 
indenting and folding along OH, OK. The three triangles Hk'B v 
bJI/B.,, bNB., are also cut in paper, with margins along their bases for 
attachment to the model by glue or paper-fasteners. 

Case II Let the given sides be a, />', and the included 
angle C. 

First draw the rabatted sides B x OC, COA equal to 
a , b. Through B x draw a perpendicular to OC, intersect- 
ing the latter in M. Set off the angle bMB 2 equal to C\ 
and make MB 2 = MB V Through B draw Bb perpen- 
dicular to BM produced. Then b is the plan of B. 
Through b draw 1>NB 1 perpendicular to OA. We thus 
find the side c, and the angles A" and B are readily 
determined by rabatments. 

If we are given two angles and an adjacent side this can 
be at once reduced to the above by means of the polar 
triangle. Or the problem can be easily worked directly. 



XIX 



MISCELLANEOUS PROBLEMS 



469 




Case III. Let a, b , A be the given sides and angle. 

First draw B\OC, CO A the development of the two 
given sides or faces. 

Next set out the elliptic arc Ab, of indefinite extent, 
which shall be the projection of the arc AB X when the 
latter is turned about OA until the face OAB is inclined 
at the given angle A to the plane OAC. 

Then through B x on the side a draw a perpendicular 
B X M to OC to intersect the elliptic arc in two points b, />. 

The plan of B is thus found, and the solution can now 
be completed as in the preceding cases. There are two 
solutions, so that Case III. is ambiguous. 

Note 3. The drawing of the elliptic arc may be avoided and ihe 
points of intersection found by the method of Prob. 95. 

The case where two angles and a side opposite to one 
of them are given reduces at once to the present case by 
means of the polar triangle. 



47o PRACTICAL SOLID GEOMETRY chap. 

364. Problem. It is required to fit a cylindrical shell 
eccentrically on a hemispherical dome, as shown in the 
figure. Draw the elevation of the intersection of the 
surfaces, and develop the plate for the cylinder. 

As the development is required, take equidistant gener- 
ators on the cylinder, and find the points where these meet 
the sphere. 

Through c, a draw the diameter of the plan of the cylinder ; 
divide the semicircles each into the same number (six) of 
equal parts, figuring the points as shown. 

Turn c$, c$ into the position c$ x parallel to xy; project 
from 3j to 3 X '; draw the horizontal line through 3/ to meet 
the projectors from 3, 3 in 3', 3'. 

Repeat the construction for the other points, and draw a 
fair curve through the points 1', 2', . . . thus found. This 
is the required elevation of the intersection. 

The development. On any line set off HAT, MK each 
equal to \ the circumference of the cylinder, Prob. 113. 
Divide HK into six equal parts. Erect perpendiculars of 
lengths M^ 1 = m'^', J\ r 2 1 = 112', . . . Draw a fair curve 
through the points o,, 1,, 2 l5 . . . as shown. 

The development of a symmetrical half of the cylindrical 
shell plate is thus obtained. 

365. Problem. To develop the surface of a sphere ap- 
proximately (a) in zones ; (b) in lunes. 

The centre of the sphere is taken in XY, and the vertical diameter 
is the polar axis. 

A zone is the surface included between two planes, perpendicular to 
the axis ; as between two circles of latitude. A lime is the surface in- 
cluded between two planes meeting along the axis ; as between two 
semicircles of longitude. 

(a) Divide the arc ric into three (or other number) of 
equal parts. Draw da, b'b' . Join b'd, b'd and produce to 
meet in v. Then the frustum AB of the cone VB coin- 
cides approximately with the zone AB. 

Develop this frustum. One half only is shown. Arc 
b'B 1 = semicircle bsb. Prob. 113. 



XIX 



MISCELLANEOUS PROBLEMS 



471 



rrV ii' a' 




~or^ 




>^ 




H 


N\ 


\M \" 


I4 6 t 



X 



The other zones must be similarly developed, 
(b) Make // = a'b'. Join ot, ot. Then tot is the plan of 
a lune of T V the surface. 



Set off oN x arc ric 



(Prob. 113). 



Divide N x into 



three equal parts by R X R V S X S V making R X R X rr, S X S X = 
ss. Draw the curves t'S v R X N X as shown. 

This is the development of the upper half of the lune. 



Examples. 1. Develop approximately the surface of a hemi- 
sphere (a) in four zones ; (b) in sixteen Junes. 

2. Two copper pipes 10" diameter, with their axes at right 
angles in one plane, are connected by an elbow pipe in the 
form of a quarter annulus, the mean radius of which is 10''. 
Develop approximately the plates for the elbow (a) in twelve 
zones ; (b) in four equal lunes. Scale x 



472 PRACTICAL SOLID GEOMETRY chap. 



368. Problem. The shape of the surface of a piece 
of ground is given by contour lines indexed in feet, and a 
horizontal scale of feet. The indexed plan of a flat sur- 
face ABCD is also given. The plot is formed partly by 
cutting and partly by embankment, the slope of the former 
being 45 and of the latter 40. It is required to draw 
the contours and boundary of the completed earthwork. 

The contours or lines of level on any inclined plane 
surface are straight lines perpendicular to the line of slope. 
Their distance apart in plan for any difference of level may 
be found by the following simple construction : 

Draw two intersecting lines, one vertical, the other 
inclined at the angle of slope, say 40^'. From their inter- 
section P set off PQ vertically to represent any difference 
in level, and draw QR horizontally to meet the inclined 
line in K. Then QR is the distance between the contour 
lines in plan. 

When, as in the above cutting, the slope is 45 , we have 
QR = PQ, or the horizontal distance is equal to the vertical 
distance ; the construction is not then required. 

We shall begin the solution by determining the contours 
at 5-feet intervals on the cutting which springs from AB. 
With a. A2 , b zx as centres, draw circles respectively of 3-feet 
and 4-feet radius to scale. The common tangent to these 
circles is the 3 5-feet contour on the cutting. The other 
contours are then drawn parallel to this line at distances of 
5 feet. A curve drawn through the points marked 1, where 
these contours intersect the given contours at the same 
levels, gives the upper boundary of the cutting on this side. 

Next consider the cutting which rises from AD. 

With ^/., 9 as centre, radius 6 feet to scale, describe a 
circle ; the tangent common to this circle and to the one 
with a. i0 as centre is the 35-feet contour on this cutting. 
The remaining contours are drawn parallel to the one 
just found at intervals of 5 feet. Their points of intersec- 
tion with the given contours are marked 2 in the figure. 
The freehand curve through the points 2 is drawn, and 



XXI 



MISCELLANEOUS PROBLEMS 



473 



10 10 ZO 50 

uilini 




35 



30 



X 

25 20 



intersects the first curve in /and the edge of the plot in e. 
These points represent the highest and lowest points of the 
cutting on this side. 

AF is the line of intersection of the two cuttings. At 
the point E the embankment begins. 

The student should now be able to complete the problem 
without further detailed description. 

For the embankments, which slope at 40 , the radii of the 
circles and the distance apart of the contours are determined 
by the horizontal lines of the scale PQR, as explained at 
the beginning. 

The line gh is the contour 30 across the plot ; BG 
and DH are each one-third the length of the plot. 

This is a good example of the use of figured plans. 



474 PRACTICAL SOLID GEOMETRY chap. 

367. Problem. A given parallelogram pqrs is the 
projection of a certain square, determine the length of the 
side and the inclination of the plane of the square. 

Draw the bisectors^ and r 

By Prob. 87 determine aa v bb v the major and minor 
axes of the ellipse of which jj v ii x are conjugate diameters. 

This ellipse will be the projection of the circle inscribed 
in the required square, and the major axis is the projec- 
tion of that diameter of the circle which is horizontal. 
Thus aa x is the length of the side of the square. 

Draw xy perpendicular to aa v and from b draw a 
projector to meet, in b', an arc with c as centre and ca as 
radius. Then the angle b'c'x is the required inclination of 
the plane of the square. 

The edge elevation of the square will be parallel to c'b', 
and we can draw it when we know the height of some 
point connected with the square. 

The rabatment of the square about AA X into a horizon- 
tal position is easily found if required. 

368. Problem. A given triangle ahc is the projection 
of a triangle ABC, the latter being similar to a given 
triangle ABC; it is required to find the actual size of 
ABC, and the inclination of its plane. 

On any side, say B'C, of the given triangle A'B'C 
describe a square ; join A' to one corner P' of this square, 
and meeting B'C in R' . 

In the corresponding side be obtain r such that 
br:rc = B'R':R'C. 

Join ar and produce to p such that 
ar : rp = A'R' : R'F'. 

Join bp and draw cq, pq parallel to bp, be. 

Then beqp is the plan of the square attached to BC\ the 
triangle ABC and the square BCQP lying in the same plane. 

Determine as in the preceding problem the inclination 
of this plane, then by a rabatment obtain the actual size of 
the triangle ABC 



XIX 



MISCELLANEOUS PROBLEMS 



475 





c I 




$ 368 



Note I. The points r and p may readily be obtained by describing 
a square on be instead of B'C, and constructing a triangle on be of the 
same shape as A' B'C. 

Note 2. The student should be able to extend the construction so 
as to solve the following more general problem : 

The plaits a, b, e are given of three known points A, B, C connected 
with any plane or solid figure ; complete the plan of the figure. 



476 PRACTICAL SOLID GEOMETRY chap. 

369. Problem. To project a helix of given diameter 
and pitch. 

Definitions. A helix is traced on a cylinder when a point travels 
round the surface at the same time advancing axially, the ratio of the 
two speeds being constant. The advance per revolution is the pitch. 

If a right-angled triangle ABC having the base AC= circumference, 
and the height Ci? = pitch, were made in paper, and wrapped round the 
cylinder, bringing the points A and C together, the hypothenuse would 
become a helix. The base angle A is the pitch angle of the helix. 

To obtain the projection, we proceed as in Prob. 130 for the sine 
curve. The latter may be regarded as the projection of a helix. 

Draw the semicircle, centre /, of the given diameter. 
Project cc of length equal to the given pitch. 

Divide the semicircle into six (or other number) of equal 
parts from o to 6' ; and divide cc into double the number 
by the perpendicular lines o to 12. 

Project from o', i', . . . on the lines 0,1,... Then 
draw the helical curve through the points as shown. 

Note. A second turn of the helix, with the projection of a helical 
spring of circular section, is shown. This is determined as the envelope of 
the projection of a sphere which maybe assumed to generate the surface. 

370. Problem. To project a screw thread and a spiral 
spring of square section. 

Definition. If the generating point above be replaced by a line, 
the helix becomes a helical surface. The line or figure tracing the 
helical surface may be conceived as attached to a screw which turns in 
a fixed nut. 

A line perpendicular to the axis, shown in the various 
positions, 00, 11, 22 . . . generates one of the helical 
sides of the screw thread. The spiral spring may be re- 
garded as having been traced by the square s. 

The pitch cc is the same as before, and the divisions o 
to 1 2 are again made use of in projecting the screw. 

371. Problem. To project a right-handed V threaded 
screw and a longitudinal section of the nut. 

After what has been said above, the construction should 
be clear from the figure, without description. 

The screw is single threaded, and the pitch pp. 

Note that a line such as act is perpendicular to the axis, since in a 
half-turn the thread advances half the pitch. 



XIX 



MISCELLANEOUS PROBLEMS 



477 








0' 


I\ 








r 


r^o 1 


12- 


i 2 ' 


-J- 1 




\j 


f 




4' 



478 PRACTICAL SOLID GEOMETRY chap. 

372. Problem. A ruled surface is generated by the 
tangent moving along the given vertical helix ABCD. 
(a) Draw the horizontal trace of the surface, (b) Obtain 
the elevation of the section made by the given vertical 
plane LM. (c) Show the envelope of the tangent in plan 
and elevation. 

The point A is on the ground, and aoc is perpendicular 
to xy. 

Draw the tangent at c, and by Prob. 113 set off ck = arc 
cb. Make ch = twice ck = rba ha\( circumference. Pro- 
ject from h to //. Join and produce h'c . 

Then HC is the tangent to the helix at C. See Prob. 369. The 
base angle a of the right-angled triangle AIIC is the pitch angle of the 
curve, and all the tangents are inclined at a to the ground. If the 
triangle AHC were wrapped round the cylinder containing the helix, 
the hypothenuse (produced) would generate the required surface, and 
the locus of H would be the horizontal trace of the surface. The latter 
is thus seen to be the involute of the plan abed of the cylinder ; this 
could be set out as in Prob. 125, or as follows : 

(a) Draw the tangent at any point d in plan. Project 
from d to d\ and draw the horizontal d'f to meet h'c in 
f. Draw the vertical _/; Mark off de equal to gk\ Pro- 
ject from e to e and join d'e. 

Then DE is the tangent to the helix at D, and E is its 
horizontal trace. Find similarly other points in the required 
trace of the surface, and draw the curve ahe through them. 

(b) Obtain the points where the plane ZJ/cuts the tan- 
gents like DE previously found. Or proceed thus : 

Select any point p in Im. Draw the two tangents qp, 
np. Draw the triangle with base angle a, and along its base 
set off H y N v H X Q X equal to pu, pq. Erect the perpendicu- 
lars dV l P v Q X P V and mark off their lengths as heights above 
xy on the projector from p, thus obtaining/', p'. 

Determine other points similar to p', p', and draw the 
required elevation of the section through them as shown. 

(c) The envelope in plan is seen to be the circle abed; 
and in elevation is the curve a'b'c'd ' . 



XIX 



MISCELLANEOUS PROBLEMS 



479 




Thus the envelope is the helix itself. The surface has 
an edge, which coincides with 
the helix. 



The figure shows how a model of 
this surface may be made. 

Set out the symmetrical curves 
on cardboard, and make holes at 
the points determined by a series of 
equiangular tangents. Indent and 
fold along the lines shown, and 
secure by paper-fasteners through 
A A, BB. 

Use twine for the generators, 
laced through the holes. 

Take the dimensions of Ex. 27, p, 



B 


I C\ 6 J 
to KJj? 


A 




pitch 




B 


f -?o 


A 


"V-p^ 



481. 



480 PRACTICAL SOLID GEOMETRY CHAR 

373. Examples on Problems 359 to 372. 

1. Draw the plan of a regular dodecahedron of i" edge (a) when 

a face is horizontal ; (b) when a diagonal of the solid is 
vertical. Measure the length of the diagonal, the distance 
between two parallel faces, and the distance between two 
parallel edges. 

2. Draw the plan of a regular icosahedron of i" edge (a) when a 

diagonal of the solid is vertical ; (/>) when a face is horizontal. 
Measure the length of the diagonal, the distance between two 
parallel faces, and the distance between two parallel edges. 

3. Determine the dihedral angles between the faces of a regular 

dodecahedron and icosahedron. Find also the angles sub- 
tended by the sides at the centres of the solids. 

4. Determine the spheres inscribed in, and circumscribing, a regu- 

lar tetrahedron, octahedron, and a cube, each of 2" edge. 

5. Determine the spheres inscribed in, and circumscribing, the solids 

of Exs. 1 and 2. 

6. Find the sizes of the five regular polyhedra inscribed in, and 

circumscribing, a sphere of 2" diameter. 

7. The lengths of the six edges of an irregular tetrahedron ABCD 

are respectively BC = 3", CA = 2", AB = 2\" , AD = 2%", BD = 
2J?", CD=2\"- Draw the plan of the solid when the face 
ABC rests on the ground, and index the plan of D. 

8. Determine the indexed plans of the spheres which are inscribed 

in, and which circumscribe, the irregular tetrahedron of Ex. 7. 
Solve the following three spherical triangles : 

9. Given a = 47, 3 = 55, c = 4l. 
Ans. A" = 62, B = 81. 5 , C = 52-4. 

10. Given ff = 75, 6 = 59, C = 5o. 
Am. A = g7.s, B = 6i.6, ^ = 48.3. 

11. Given a = 55 , b a = 8o, A" = 45 '. 

Aits. c* = 6f, B =.$S.2, C =i27\ 4 ; 
or _ ^==39.1, ^=i2i.8, C = 33- 

12. Taking the answers in the above three examples as data, solve 

the four triangles. 

13. The latitude and longitude of Rome are respectively 41.9 N. 

and 12.5 E., and of St. Petersburg 6o.o N. and 30.3 E. : 
find the angle subtended by the two places at the centre of 
the earth ; find also the direction at each place of the great 
circle connecting the two. Ans. 25.05, 140.8; 21. 2. 

14. Two copper pipes each 10" diameter meet at right angles to 
form an elbow joint. Draw the development of the plate near 
the joint for one of the pipes. Scale V'. 

15. A horizontal range of sheet metal piping, 6"diameter, is required 
to have a horizontal 4" branch fitted at an angle of 6o, the 



xix MISCELLANEOUS PROBLEMS 481 

upper surfaces of both pipes being at the same level. Draw 
the development of the joint for both pipes. Scale ^. 

16. Work Prob. 365, the diameter of the sphere being 4". Take 
four zones in the hemisphere and develop them all. Let the 
angle of a lune be 22^. 

17. A square with one corner on the ground projects into a par- 
allelogram, with adjacent sides 3" and 2", and included angle 6o\ 
Determine the side of the square, the trace and the inclination 
of its plane, and the heights of its other corners. 

18. Draw a parallelogram abed, taking ab = . 7", ad=^', 6ad=5$. 
This figure is the projection of a rectangle whose sides AD, 
AB are in the ratio of 3 to 2. Determine the sides and the 
inclination of the plane of the rectangle. 

19. An equilateral triangle of 2" side is the projection of a right- 
angled triangle whose three sides are in the ratios 3:4:5; 
determine the size and angular position of this latter triangle. 

20. A right-angled triangle whose three sides are i\" , 2", 2j" is 
the projection of an equilateral triangle ; determine the size 
and angular position of the latter. 

21. Draw a triangle oab making oa = 2", ob 1. 7", ab = 1. 3". The 
point is the plan of the centre, and the line ab that of one 
side of a regular octagon ; complete the plan of the figure. 

Special solution. Draw any regular octagon ABC . . . centre O. 
Join A C to intersect OB in M. In the given triangle take ;// 
in ob such that om : tub = OM ': RIB ; join am and produce to 
c, making mc am. The student should be able to complete 
the solution. 

22. In a regular tetrahedron of 3" edge take points P, Q, R on three 
edges distant respectively 0.6", 1.5', and 2.2" from one corner. 
Draw the plan of the solid when the triangle PQR projects into 
an equilateral triangle. 

23. Draw a triangle abe, have abT,", ae = 2", and 6e=l^"; ab is 
the plan of one edge of a regular tetrahedron, and e that of the 
middle point of the edge opposite to AB. Complete the plan 
of the tetrahedron. 

24. Work Prob. 369, having given: diameter=3"; pitch = 4"; 
diameter of section of springs 1". 

25. Work Prob. 370, having given: diameters = 4" and 3" ; pitch 
4" ; width of thread = -|". 

26. Work Prob. 371, having given : diameter outside = 4"; angle 
of thread = 6o ; pitch = J" ; length of nut = 4". 

27. Work Prob. 372, having given : diameter of helix = 1"; pitch 
= 2^" ; lm makes 15 with xy and touches the circular plan ac. 
Make a model of this surface as described in the problem, using 
stout cardboard. 

2 I 



482 PRACTICAL SOLID GEOMETRY chap. 

374. Miscellaneous Examples. 

1. A regular tetrahedron of 2" side is inscribed in a sphere. Draw 

the elevation of the solids, the circumscribed sphere resting on 
the horizontal plane, one face of the tetrahedron horizontal, and 
one edge of that face inclined at 20 to the vertical plane. 

f (i897) 

2. Draw a triangle oab, having ab = .gz" and oa = ob = .$8". Take 

as the plan of the centre of a sphere, radius ", and a, b as 
the plans of two points on its upper surface. Determine the 
plan of a regular triangular pyramid, circumscribing the sphere, 
and having two of its sloping faces touching the sphere at A and 
B respectively. 

3. An octahedron of 2" edge stands on the horizontal plane. Draw 

its plan and also that of its inscribed sphere. (1884) 

4. Draw the plan of a cube inscribed in a sphere of 3" diameter 

resting on the horizontal plane, one corner of the cube to be at 
a height of 2.75" above the horizontal plane. (1878) 

*5. Eig. (a). Determine the centre and radius of the sphere circum- 
scribing the irregular pyramid ABCD. Unit = 0.1". (1882) 

" ;; "6. Fig. (e). The three given parallel lines are the plans of portions 
of the edges of a triangular prism. Determine the plan of a 
sphere inscribed in the prism, and having 16 as the index of 
its centre. Unit = 0.1". (1887) 

*7. Fig. (d). The arc be is part of the plan of the base o. a right 
cone standing on the horizontal plane ; v u is the plan of the 
vertex. The portion ABCD is to be covered with paper. 
Determine the shape to which the paper must be cut. Unit 
= 0.1". (1888) 

*8. Fig. (b). The end elevation and a portion of the plan of two 
adjacent ridge roofs is given. Determine in plan the shortest 
distance measured on the roof surface from A to B. (1887) 

*9. Fig. (e). Determine the shape of the plates used to form the 
elbow pipe with bell-mouth shown. Draw the figure four 
times the size shown. (1887) 

10. Fig. (f). Draw the development of the funnel shown, so as 
to give the shapes of the plates from which it is made. Draw 
the figure four times the size shown. 

11. Draw a rectangle abed (ab = cd 3^" ; be = da=i^"); ah and cd 
are the plans of the diagonals of two opposite faces of a cube ; 
be and da are the plans of two edges. Complete the plan of the 
cube. (1880) 

12. Two equal right cones, height i|", diameter of base 2-3", have 
a common vertex. One rolls upon the other which stands on 
the ground. Determine the locus of a point on the circum- 
ference of the base of the rolling cone. (Honours, 1SS6) 



XIX 



MISCELLANEOUS PROBLEMS 



483 




484 PRACTICAL SOLID GEOMETRY chap. 

*12. Fig. (a). The area abed on a hill side, the traces of the plane 
of which are given, is to be levelled ; the side slopes are I in 1 . 
Complete the plan of the excavation. (1885) 

*13. Fig. (/). The plan and section of an embankment are given ; 
the lines aa, bb represent the sides of a road cut through it. 
The slopes of the sides of the cutting are 35. Complete the 
plan. ' (1884) 

14. On a right circular cylinder of i|" diameter, a helix of 3" 
pitch is traced. Draw the elevation of one turn of the helix : 
draw also the plans and elevations of tangents to the helix at 1 2 
equidistant points, and determine the points in which these 
tangents meet the horizontal plane through the lowest point of 
the helix. Axis of cylinder vertical. (1890) 

15. A spiral spring, axis vertical, is of the form of a square screw 
thread. Side of square V : external diameter on plan 3 " : 
pitch 2|". Draw the elevation of one complete turn of the 
spring. 11877) 

16. A right cone, height 4", diameter of base 3", stands on the 
horizontal plane. A point starting from the base of the cone 
moves round its surface at a uniform speed, and rises half the 
height of the cone in turning round to cut the generatrix from 
which it started. Draw the plan and an elevation of the curve 
traced by the point. (1SS6) 

"*17. Fig. [e). A twisted surface of revolution is generated by a line 
rft, a'c', revolving round a vertical axis 0, 00, to which it is 
rigidly fixed by the horizontal line ob (in elevation />'). Draw 
in plan and elevation the generating line when it has revolved 
round the axis so as to pass in plan through/. (1894) 

*18. Fig. (<). Two lines are given by their figured plans ab, cd. A 
surface is generated by a line moving parallel to the horizontal 
plane and always meeting both the given lines. Determine the 
true form of the section of this surface by a vertical plane LM. 
The section to be carried up to a height of 3" above the ground. 
AYhat is this surface termed, and can it be developed? (1879) 

19. Fig. (</). Two views of a crank are given. Determine the 
proper shape of the dotted curve in the right-hand view. 

20. A line 3" long moves at a uniform speed round a vertical axis, by 
which it is bisected, and to which it is always at right angles. At 
the same time it moves along the axis a distance of 1 |" for every 
complete revolution, thus generating a helical surface. Make a 
section of this surface by a plane parallel to and J" from the fixed 
axis, the section to show two revolutions of the surface. (18S9) 

21. Fig. (/"). A surface of revolution (a capstan) is shown in 
elevation ; determine a sectional elevation on a vertical plane 
i" from the axis. 



XIX 



MISCELLANEOUS PROBLEMS 



4S5 



Copy tfie figures double size- . 




IV 
3 




i0h*& 


- ^ 


~~ 7 u 





C o 


\o' 


'e' 


r -7 


Xb* 


(e) 


a'/ 


\o' 




U 


CI 




e 



<0 p 




SECTION III 
GRAPHICS 

CHAPTER XX 

GRAPHIC ARITHMETIC 

375. Graphic representation of magnitude. In ordinary 
arithmetic we are concerned with numbers, and with numeri- 
cal calculations. The former may be pure numbers, or they 
may be concrete, that is, may express the magnitudes of 
quantities of some specified kind. In the latter case the 
number measures the magnitude by telling us how many 
times the quantity contains an arbitrarily chosen unit of the 
same kind and of known size. 

In graphic arithmetic, numbers and magnitudes are re- 
presented by the lengths of straight lines, set out or 
measured to scale ; and calculations are made by drawing. 

Thus let it be required to represent the number 7.5. 

We may first draw any line, setting off any convenient 
length on it to represent unity, and then on the same or 
any other line mark off a segment 7.5 units long. 

Or instead of drawing the unit line, we may set out or 
specify some convenient scale, say J" to 1 unit, and using 
this scale mark off the segment 7.5 units long. 

If the number had been concrete, say 7.5 tons, then the 
unit line would have been labelled 1 ton, or the scale would 
have specified |" to 1 ton. 



chap, xx GRAPHIC ARITHMETIC 4S7 



II the number to be represented is comparatively large or small, as 
for example 750 or .075, then we may replace the unit line by a line 
which shall represent some convenient known multiple, such as 100 
units, or submultiple like .01 unit. The scales are also correspond- 
ingly modified ; thus we may take as suitable scales for plotting these 
numbers ^" to 100 units, or V to .01 unit. 

The student should refer to Probs. 7 to 1 1 on scales. Also to Art. 
5 for the description of an engine-divided decimal scale, by the use of 
which we are enabled to pass from the numerical to the graphical 
system or vice versa. 

376. Addition and subtraction. Lines to be added or 
subtracted must represent quantities of like kind, and be 
drawn all to the same scale. 

The process of addition consists in drawing the lines in 
position, end to end, all in one direction, so as to form a 
straight line equal to the sum. If lines are to be sub- 
tracted, they must be set off in the opposite direction. 

A' ' o-A-->* B { 

\ 1 ! k C 4 

Cl , j '^._ D ._._^ 

D\ 1 1 ! 1 Ljy 

PR S Q 

Thus in the figure OP= A ; OQ=A+B; OR = A + 
B- C; OS=A + B- C + D 

Note 1. Observe that the sum OS would be the same if the 'terms 
were added in any other order, say in the order D + B + A C. 

Note 2. The summation is most readily effected by applying the 
edge of a strip of paper in succession to the lines A, B, C, D, and 
marking the points O, P, Q, R, S with a sharp pencil. 

377. Multiplication and division. The product of two 
lines, like that of two numbers, means that one is to be re- 
peated as often as there are units in the other. That is 

Unit line : one line : : other line : product. 
And in division we have the proportion 

Divisor : dividend : : unit : quotient. 
Thus graphic arithmetic consists mainly of finding fourth propor- 
tionals to given lines. The results may be represented graphically, or 
expressed numerically by measuring on the unit scale. 



4 S8 GRAPHICS chap. 

378. Problem. Having given two lines A, B, and the 
unit line S ( = V), to find the product of A and B. 

Let AxB=X=Xx i=XxS. 

A X 

From which =, =tt5 ox S : A : : B : X. 

o h 

We thus find X as a fourth proportional to .S', A, B as follows. 

Draw two lines intetsecting at O. 

Along one set off OA equal to A. Along the other set 
off OB and OS equal to B and S. 

Join SA, and draw BX parallel to SA. 

Then OX represents the required product graphically. 

Measuring OX on the scale of S, or -h", to one unit, the 
product is 2.94. 

379. Problem. Having given two lines A, B, and the 
line S ( = J"); to find the quotient A -=- B. 

Let 4- = X = -= ~ . Or B : A : : S : X. 

B 1 6 

So X is a fourth proportional to B, A, S, and may be found thus : 

Draw two lines intersecting at O. 

Along one set off OA equal to A. Along the other set 
off OB and OS equal to B and S. 

Join BA, and draw SX parallel to BA. 

Then OX represents the required quotient graphically. 

Measuring OX on the scale of .S", or |", to one unit, the 
quotient is 2.25. 

Note. With a different unit A and B would represent other numbers, 
but their quotient when measured would be the same number. 

380. Problem. Having given two lines A, B, and their 
product X, to find the unit line S. 

This problem is the converse of Prob. 377- 
In Fig. 378 set off along one axis OX and OA equal to 
X and A. Along the other set off OB equal to B. 
Join XB, and draw AS parallel to XB. 
Then OS is the required unit length. 

381. Problem. Having given two lines A and B, and 
X the quotient A-fB, to find the unit line S. 

This problem is the converse of Prob. 378. 



XX 



GRAPHIC ARITHMETIC 



489 



o 




378 



.K 



A 



A 
B 



1 II 1 



X 




I 







A 



B 



5\ 



n 1 



379 



X 



A 



In Fig. 379 set off along one axis OX and OA equal to 
X and A. Along the other set off OB equal to B. 
Join ^4i?, and draw XS parallel to AB. 
Then OS is the length of the required unit. 

Examples. 1. The unit line being 1.34" long, construct the unit 
scale, and mark off a length representing 2.27 units. 

2. If the line S, Fig. 379, represent unity, what numbers do the 

lines A and B represent? Ans. 3.60 ; 1.62. 

3. In Fig. 379, taking S as the unit of length, find the number of 

units of area in the rectangle, two adjacent sides of which are 
equal to A and B. Ans. 5.83. 

4. Find the line which represents |- to a unit of -J-". 

5. In Fig. 379, if S be the unit, represent graphically :: . 

A 

6. Draw the lines A and B of Fig. 379 double the length. Then 

A - B 
determine the numerical value of r rv Ans, o. ?86. 

A + B 

7. In Fig. 379, if 5 represents the product of A and B, find and 

measure the unit of length. Ans. 2.92". 

8. If the line ./, Fig. 379, represent the fraction , determine 

3-9 

and measure the unit of length. Ans. 1.02". 



490 GRAPHICS chap. 

382. Problem. Having given the lines A, B, C, . . . 
and the unit line S ( = |"), to find the continued product 
A x B x C x . . . 

The construction is a repetition of that of Prob. 37S. 

Draw two lines intersecting at O. 

Along one set off OA equal to A. Along the other set 
off OS, OB, OC, . . . equal to S, B, C, . . . 

Join SA, and draw BX 1 parallel to SA. Join SX V and 
draw CX, parallel to SX 1 ; and so on. 

Then OX x = Ax=i.s; OX 2 = A x B x C= 2.6, etc. 

383. Having given the lines A, B, C, . . . and the 

unit line S ( = I"), to find the value of - - 

B x C x . . . 

The construction is a repetition of that of Prob. 379. 

Along one axis set off OA equal to A. Along the other 
set off OS, OB, OC, . . . equal to S, B, C, . . . 

Join AB, and draw SX 1 parallel to AB. Join CX V and 
draw SX, parallel to CX t ; and so on. 
A A 

Then OX 1 = = 1.93 ; OX, = -g^-= I * 5 ' et ' 

B D 

384. To find the value ofAx-x-x . . . 

Set off the lengths A, B, C, D, E, . . . from O along 
the two axes in the manner shown in the figure. 

Join CB, and draw AX X parallel to CB. Join ED, and 
draw X X X^ parallel to ED. And so on. 

Then OX, = A x . OX, = Ax x , etc. 
1 C C E 

Note. The Roman numerals in the figures indicate the order in 
which the dotted lines between the axes are drawn. A reference such 
as II. || I. means that line II. is to be drawn parallel to line I. 

Examples. 1. In Fig. 383, if S be the unit of length, find the 
number of units of volume in the rectangular prism of which A, 
B, and Care three edges. Ans. 19.8. 

2. Determine graphically the value of 3.14 x i.7 2 ^o.67. 

Ans. 13.54. 

C E . ... 

3. In Fig. 384 determine A x -^ x , the unit being i . Ans. 7.3. 



XX 



GRAPHIC ARITHMETIC 



491 





J I || I 







X z X, 



383 



A 











384 



Mr 



C D X 9 



-A 
-If 



JL II T 

w II m 



492 GRAPHICS chaf. 

A C 

385. Problem. To find the value of + - + . . . 

B I) 

The method of solution is to reduce the fractions to a common 

denominator K=k units, where k is any known number (say 3). 

A A' C A'., 
Thus let - B = -^; ^ = ^; . . . 

A C 

Then X t = Xx -= ; A 2 =A'x-vr; . . . and these values may be 

found as in Prob. 383. 

Set off OK= K, say f ", OA A, . . . along any con- 
venient axes in the manner shown. 

Determine X v A'. ... by drawing the pairs of parallel 

lines between the axes in the order indicated by the Roman 

numerals. 

A C 
Then + + . . . = OX, + OXc, + . . . measured on 
Jy JJ 

the unit scale and divided by k ; or measured directly on 

the scale on which OK (f") is the unit. 

386. Problem. To find the value of A\B + CD + . . . 

The series of products are first reduced to any convenient common 
base A' (|") representing k units, k being a known number (say 3). 
Thus let A-B = X 1 -JsTj C-D=X^K; .... 

Tl V A ' B V C ' D 

Then X 1 = ; X t = -g- ; . . . 

These values may be found as in Prob. 383. 

Make the construction which is sufficiently indicated by 
the notation of the figure. 

Then AS + C-D + . . . = OX x + OX 2 + . . . measured 
on the unit scale (\") and multiplied by k (3) ; or measured 
directly on the scale on which OK (f ") = U 2 (or 9) units. 

387. Probl*em. To find the number of units of area in 
a given irregular polygon, having given the unit of length 
S. 

First reduce the given polygon by Prob. 30 to an equivalent tri- 
angle, then find half the product of the base and altitude of the triangle. 

In the figure OA = altitude ; OB~\ base ; OS= unit length. Then 
OX measured on the unit scale gives the required area. 



XX 



GRAPHIC ARITHMETIC 



493 




X, X ? A 








w 



a 



C AX 



R II I 
W || IE 



X, 



A 

B 

< C 
J? 




387 b 




Hi 



r bo 



A X. 



494 GRAPHICS chap. 

388. Involution and evolution. Involution is the pro- 
cess of raising a number to any integral power. Thus A n 
(read A to the n\h power) stands for the continued product 
A x A x . . . , where the factor A occurs n times. n is 
called the index of the power, or simply the index. The 
value of A n can be found by Prob. 381, putting B, C, . . . 
each equal to A. It may also be found by the general con- 
struction of Prob. 394. In this case, however, the special 
method given in the next problem is preferable. 

Evolution, or the extraction of roots, is the inverse of 

involution ; thus v A, the nth. root of A, denotes a quantity 
which being raised to the nth. power gives A as the result. 
The general graphical solution requires the use of a logarith- 
mic curve or spiral, and is given in Prob. 394. The extrac- 
tion of the square root, and of the fourth, eighth, sixteenth 
. . . roots may however be effected by simple geometry. 
See Probs. 390 and 391. 

389. Problem. Having given a line A, and the unit 
line S, to find the integral powers of A, viz. A' 2 , A 3 , . 
Also to find the integral powers of the reciprocal of A, viz. 
1-7- A, 1-A 2 , . . . 

Draw two perpendicular axes intersecting at O. 

Along them set off OA and OS equal to A and S. 

Join SA, and draw the lines AX, XX V . . . SY, YY V 
. . . alternately perpendicular and parallel to SA, as shown 
in the figure. 

Then OX=A 2 ; OX l = A 3 ; . . 

And OY=~; OY.=~; . . . 
A' A % 

These results follow from the similarity of all the tri- 
angles formed by the axes and the dotted lines. 

Note. If OS had been made equal to s units on the scale for A, 
then the scale for measuring OX ox A' 1 would have been OS to s' 2 units; 
and for OX 1 or A 3 , OS to s 5 units ; . . . And the scales for measuring 

OY. OY,, . . . i.e. , , r , . . . would have been OS to - , -, . . . 
1 A A* s s z 

unit respectively. 



XX 



( ; RAP] IIC ARITHMETIC 



495 








X_LI 








111 


4 

1/ 


v. 


x, 




Y/ 


o X>4 


389 


\ 


nrN 


r 

Y 

l 
fa 

f 

/ 






V 


t 
/ 

/ 








X 



Examples on Problems 384 to 388. 

Q M 

1. Find the value of -", + -54--^. Ans. 4. 86. 

2. Find the value of P'Q+Q-R + M'JV, the unit being 1". 

Ans. 1 5. 8. 

N 

3. Find the value of + Q'M, the unit being R. Ans, 5.49. 

4. If P be the unit of length, find the values of Q 2 , Q :i , , , , --. 

Ans. 5.35, 12.37, .432, .187, .0808. 

5. Find the line which represents the cube of a line 1.37" long to a 

unit of 1". Ans. 2.57". 

6. If the line N represent the square of the number represented by 

the line R, find the length of the unit. Ans. 1.3". 

7. A line 2.9" long represents the sum of the areas of an equilateral 

triangle and square, each of 1" side. Determine the unit of 
length. Ans. .494". 

8. Draw a curve which will give the squares of all quantities from o 

to 6, taking J" as the unit. 



496 GRAPHICS chap. 

390. Problem. To find the square root of a given line 
A, having given the unit line S. 

Let A = J A ; then A" 2 = A=Axi=AxS. 
That is -rr= ''; A : X : : X : S. 

Thus the required square root X is a mean proportional between A 
and S. See Prob. 27. 

Draw a straight line, and from any point O in it mark off 
in opposite directions OA and OS equal to A and .5'. 

Bisect SA in C, and with centre C draw the semicircle on 
SA. Draw OX perpendicular to SA. 

Then <9X ( - 1.54) = sj OA x OS= J~aVi= J At. 

If S be not a unit line, OX represents the square root of A x S. 

Note 1. If the construction were repeated on OX, the 4th root 
OA would be obtained. And by continued repetition the 8th, 1 6th, 
. . . , 2th roots might be found. See next problem. 

Note 2. If A were a large or small quantity compared with the 
unit .S", an "ill-conditioned" construction might be avoided by making 
OS equal to s units, where s is any convenient known number. In 
which case the square root OX would require to be measured on the 

scale of OS to *Js units ; the 4th root on the scale of OS to *J s units ; 
and so on. See next problem. 

391. Problem. To find the eighth root of 614. 

To any convenient scale set off OA equal to 614 units, 
and OS equal to say 2 s , that is 256 units. 

Draw OX perpendicular to SA, and describe the semi- 
circles in the order figured, making OB, OC equal to OX, 
OY. 

Then ^614= OZ measured on the scale on which OS 
= v 256 = 2 units. 

392. Problem. To draw a figure which shall give the 
square roots of the first n natural numbers. 

Draw OA of unit length on any convenient scale. 
Draw AB perpendicular and equal to OA. Join OB. 
Draw BC perpendicular to OB and equal to OA. Join 
OC. And so on. 

Then OB= J 2; OC= Jz; . . . (Euc. I. 47-) 



XX 



GRAPHIC ARITHMETIC 



497 




s o c 



A 



S 



390 



OB = 


CJC 


, OC = OY 


X 


-__ 


, ' 


Y 




/ * 


z~- 


/ 






1 _ 




! ': i 



S O CB A. 

391 




392 



Examples. 1. Determine graphically \/s.2, ^273, and 
\l 0.035. .4ns. 2.28; 16.5; .187. 

2. Determine ^165, v/o.165, .3I4 3 > and .031 4 3 . Am. 3.58; 

.637 ; .031 ; .000031. 

3. A line 2.6" long represents 4^ units. Obtain lines which repre- 

sent 1. 3, V, and 2 ^"6 units. Am. .75"; 1.93"; 2.83". 

4. On page 495, taking P as the unit line, determine the line which 

represents JiV-^Q. Ans. .694". 

5. If the line N on page 495 represent the square root of the line 

Q, determine the unit line. Ans. 6.06". 

6. If a line i\" long represent J 2, determine a line representing 

J J. Ans. 2.16". 5 

7. Taking V as unit, obtain lines representing 15, Ji$, ^ 

Ans. 3-75"; -97"; .323"- VIS - 



8. Taking the lines/ 5 and Q on page 495, determine JQ^ + P 2 , 

JQ 2 - P\ and the ratio sj ? ~ ]. Unit = o. 5". 

Hint. Make use of Euclid I. 47. Ans. 4.03; 3.33; 1.21. 

2 K 



498 GRAPHICS chap. 

393. Examples worked out. 

1. in and n are two given lines, determine s/mln, the unit bsing 

0.25". 

Set off OS, Fig. 1 (a), equal to the unit, 0.25". 

In the opposite direction set off OFF, M equal to tn and ;/, 
and describe semicircles on SN and S3I as diameters. 

Draw 0M X at right angles to OM, then 0M X is dm, and 

07V X is \fn. 

Set off these lengths and OS ( = j") along two axes on (/>) as 
shown. Join I\I X N X , and draw SX parallel to M X N X . 

Then OX represents \'m/n, and measured on the j" scale 

is 1.5. 

s/t'AB-CD 

2. Find a line which represents p= . Unit=\\ . (1896) 

_ v 7 3 # 

sJ^-AB-CD^r V3 is the same as \UaB-CD. Set off 
AB and divide it into three equal parts as shown in Fig. 2 : 
make BE equal to one of the parts ; then AE \AB. 

Make FF= CD ; on AF describe a semicircle, and erect 
the perpendicular EG ; then EG represents sj%CD'AB. 

3. If ^^.^R, find the unit of length. Fig. 3. 

This may be written 

Px Q R . R 2V 

~v Ar~ ~* ' or ' unlt ~-^ x ~r> x ~r\' 

R x JV unit P Q 

We may now proceed as in Problem 384. 

. A' 2 N 
Or, write, unit = = . y 

Then set off OP=P; and, at right angles, OR = R. 

Toin PR, and draw RA" l perpendicular to PR to meet PO. 

Then 0X X represents R 2 +P; for OP-OX l = OR 2 . 

Make OQ and OJV equal to Q and jY ; join QN, and draw 

X X X 2 parallel to QN. 

R 2 A r . 

Then 0X 2 represents , or unit line; it measures 0.76". 

4. Determine the value of (o. 2S9) 3 . Fig. 4. 

Set off OA equal to .289 on the scale ' to o. 1. 

Set off OS equal to V', i.e. . 2 or i unit. 

Join AS, and draw AX, XX X respectively perpendicular and 
parallel to AS. Then 0X X represents (.289)". 

Measuring 0X X on the scale of OS to (.2) 3 of a unit, that 
is, V' to .008 unit, or -^" to .001 unit, the answer is .024. 



XX 



GRAPHIC ARITHMETIC 



499 



17t\- 



/! 



/ 



Rgl 



/ 


(M, 




^~~~ 


/ 


s* 


1 / 

1/ 


N, 


1/ 




1/ 
1 

1 





(a) 



K/ (b) 



S/\ 



S 



If 



\ v. 



_-- G 



L 



[\ 
\ 
v 
\ 



MO X M, 
Ai 

c 



B 



J) 



Q. 



B E 
Fig. 2 



F 



s; 



p 



O NX, 



^T^ 5 



<R 









~3. 



'/ y 



Fig., 



R 



Q 



N 



A 



O 



1 



x, 



/ Fig. 4. 



500 GRAPHICS chap. 

394. Problem. Having given a line A and the unit 
line S, to find the value of A", where n is any given 
integer or fraction, positive or negative. 

By the construction of Problem 129 set out on the base 
BB any suitable logarithmic curve LL. 

Determine OS, the ordinate of unit length S. 

On OS set off OA equal to A. Draw AP, PM par- 
allel and perpendicular to BB. 

Mark off ON = n times OM; in the same direction as 
OM if n is positive, and opposite if n is negative. 

Draw JVQ, QX perpendicular and parallel to BB. 

Then OX measured on the scale of unit S gives A n . 

Note 1. The logarithmic curve might be used to perform graphic 
arithmetic in a manner exactly analogous to the way in which a table 
of logarithms "is employed in making arithmetical calculations. 

Ordinates such as PM, QN, represent numbers, and the abscissae 
OM, ON, their logarithms. 

Note 2. The logarithmic curve might be replaced by a logarithmic 
spiral, see Prob. 128. 

395. Miscellaneous Examples. 

1. Determine graphically 2.5s, i.9~?, and 0.72 -0 ' 6- Ans. 1.84 ; 
.76 ; 1.22. 

*2. If -^=C, find the unit of length. Ans. if". (1898) 



*3. Find a line which represents \t ^'A'B-r- V3. 

(1896) 

*4. Copy the figure double size. Then reduce it to an equivalent 
triangle with its base on All and its vertex at E ; and find the 
length of a line representing the area of the figure, taking for 
unit 1". O894) 

c /a*-b 2 

*5. Find the ratio , \J ^ 2 , the unit of length being 2". 

Ans. .932". Hint. Compare Ex. 8, p. 497. (1890) 

6. Determine by graphic arithmetic a line representing the con- 
tents of a rectangular solid whose dimensions are 3" x 1.75" x 
1.25". Unit = 2.5". Ans. 1.05" long. (1897) 

7. Find a line to represent x 5 when x = 2 \J 1 - \ 3. Unit= 1". 
Ans. 1.32". (1895) 

8. A line 3" long represents the sum of the areas of a pentagon, 
square, and equilateral triangle of 1" side. Determine the unit 
of length. Ans. 1.025". O891) 



XX 



GRAPHIC ARITHMETIC 



5oi 




N B 



C\ 



D 



^B 



As 
3 B> 




CHAPTER XXI 



GRAPHIC STATICS 



396. Directed quantities. In arithmetic and algebra we 
are concerned with number and magnitude. A second 
important quality possessed by many objects and phe- 
nomena, and which can be made the subject of calculation, 
is that of having a definite direction in space. Graphical 
processes are uniquely fitted to deal with this property. 

Every one is acquainted with such quantities, and would, 
for instance, distinguish between a length measured to the 
north, and the same measured, say, to the east. 

The student will readily call to mind other instances in 
which direction is associated with magnitude. He may 
think of a force of so many pounds weight, and acting along 
some definite line ; of a top spinning at a definite speed 
about an axis, perhaps inclined ; or of a field of force, 
which at any point has intensity and direction. The word 
vector is a general term, used to denote all such directed 
quantities. 

Some kinds of quantities like volume, time, mass, 
temperature, energy, are essentially non-directive ; others, 
like forces, velocities, which have direction, may yet be 
treated arithmetically as to magnitude only. But in the 
class of problems we are now about to investigate, it is a 
cardinal feature that the directions of the quantities which 
enter into the case, as well as their magnitudes, shall have 
influence on the result. 



CHAP. XXI 



GRAPHIC STATICS 



503 



J 



1 



L 







a 



Scale I " to 50 feet 



397. Graphic representation of a vector. The relative 
positions of points and their changes of position are amongst 
the simplest examples of directed quantities. Such quantities 
may evidently be represented with great convenience and 
directness on paper by directed lines. 

For example, a displacement of say 50 feet to the east 
might be exhibited thus : Let the direction to the north be 
indicated; then (1) draw a line running east and west; (2) 
on this line mark off a segment representing 50 feet to a 
suitable scale, and specify the scale ; and (3) to the segment 
affix an arrow-head pointing eastwards. In the figure above, 
oa measures 58 feet. 

The change of position is thus represented completely 
and without ambiguity, it being understood that the dis- 
placement is a horizontal one. The arrow-head indicates 
what is called the sense of the directed line ; it distinguishes 
an eastward from a westward movement. Sometimes if is 
convenient to indicate the sense in another way. The ends 
of the line are marked and named in the order or sequence 
corresponding to the beginning and end of the displacement. 
Thus oa would indicate a displacement in the sense of to 
a, or eastwards. If the ends of the line were named in the 
sequence ao, this would be understood to indicate the 
opposite or westward direction, from a towards 0. 

Example. Mortlake is 2 miles west of Putney, and Wimbledon is 
3 miles south of Putney. Plot the relative positions of these places to 
a scale of 1 inch to the mile, and measure the distance and direction 
of Wimbledon to Mortlake. 

Aus. 3.6 miles, 33.7 west of north. 



504 GRAPHICS chap. 

398. Resultant and components. Rectangular com- 
ponents. When a series of displacements are given to a 
body, the change of position represented by the straight 
line which joins the initial to the final position is called 
the resultant displacement, and the several displacements 
are called components. 

Any displacement may be imagined to be made up of 
components, and in an indefinite number of ways, just as 
we may walk from one place to another by an indefinite 
number of routes. But there is only one resultant which 
corresponds to a given set of components. 

We are said to compound a set of displacements when 
we combine them to obtain their resultant, and in the 
reverse operation we resolve a vector into components. 

As just stated, a vector can be resolved into components 
in an indefinite number of ways. For the case, however, 
of two components which are parallel to given lines, there 
is only one solution. If the two lines are perpendicular to 
each other, the components are said to be rectangular. 

Suppose a person to walk from one corner of a room 
for a distance of 10 feet, in a direction making 30 with a 
side wall. He would reach the same spot by walking 
8.67 feet along the side, and then walking 5 feet parallel 
to the end wall ; or he might first walk 5 feet along the end 
wall, and then 8*67 feet parallel to the side wall. The 
rectangular components of his change of position parallel to 
the side and end walls are respectively 8.67 and 5 feet. 

Thus the rectangular component of a vector in any direction 
is obtained by projecting the vector on a line parallel to the 
direction. This is often called simply the component in 
that direction, rectangular being understood. 

For example, the component of the given vector oa 
parallel to the given direction XX, is equal to ob, or /;///, 
obtained by drawing om, an perpendicular to XX, and ob 
parallel to XX. 

It is important that the student should be familiar with 
the idea of resolving a vector in a given direction. 



xxi GRAPHIC STATICS 505 




X i > i X 

m n 

Examples on Articles 398 and 399. 

1. A person journeys 1 mile eastwards and \ mile to the norlh-east, 

find his resultant change of position. Find also the northerly 
and easterly components of the resultant. Ans. 1.62 mile, 
19. 1 N. of E. ; .53 mile, 1.53 mile. 

2. A football is kicked a distance of 52 yards in a direction making 

35 with the side lines of the field. Find the component dis- 
placements of the ball forwards and sidewise respectively. 
Ans. 42.6 yards, 29.6 yards. 

3. Find the component of a vector 3.7 units long, in a direction 

making 53. 6 with the direction of the vector. Ans. 2.18 miles. 

4. The two rectangular components respectively southwards and 

westwards of a horizontal vector are 31.3 feet and 20.7 feet. 
Determine the magnitude and direction of the vector. 
Ans. 37.1 feet, 56.5 S. of W. 

5. A boy on a raft walks 9 feet from a point A to a point B, and 

during this lime the raft moves (without rotation) a distance of 
14 feet in a direction making 115 with AB. Find the re- 
sultant motion of the boy. Ans. 1 3. 1 feet; 76. 4 with AB. 

6. A ship at sea sails 8.7 miles through the water apparently to 

the east, but an ocean current simultaneously carries the vessel 
3.4 miles to the south-west. Find the resultant movement of 
the ship, i.e. its displacement as regards the bottom of the sea. 
Ans. 20. 9 S. of E. ; 5.88 miles. 

7. A body receives component displacements of 7.38 and 5.16 feet 

in two directions which make 7 J -3 with one another. Find the 
resultant displacement. Ans. 9.58 feet, 47 with second vector. 

8. Two places are 1. 29 miles apart in the direction north and south. 

A person journeying from one to the other first walks to the 
north- north-east, and then to the north-west. Find the lengths 
of the two stages of the journey. Ans. .987 mile, .534 mile. 

9. If a person walk between the two places of the preceding 

example in two straight paths, one of 1 mile, the other of -^ 
mile, find the directions of the paths. Ans. 20.6", 44.8". 

! 



506 GRAPHICS chap. 

399. The triangle and parallelogram vectors. Suppose 
an object to be placed on a table in a room. Let the 
position of the object be changed on the table in the 
manner represented by the directed line P, and let the 
position of the table in the room be also changed as re- 
presented by Q. (The table must remain parallel to itself.) 
It is required to find the resultant displacement of the 
object in the room. 

From any point a draw ab equal and parallel to P and 
similarly directed ; and from b draw be equal and parallel 
to Q and similarly directed ; mark ac with an arrow-head 
pointing from a to c ; then the directed line ac represents 
the resultant displacement. The triangle abc with the 
arrow-heads or their equivalent is called the triangle of 
displacements. It is a vector triangle. 

If the movement of the object on the table had pre- 
ceded the movement of the table, the path of the object in 
the room would have been similar to ab, be ; if the displace- 
ment of the table had been effected first, the path would 
have been similar to ad, dc, where abed is a parallelo- 
gram ; if the two movements had occurred simultaneously, 
each at a constant rate, beginning and ending together, the 
path would have been a straight line similar to ac. Any 
path, stepped or curved, could be effected by suitably 
timing and adjusting the two component displacements ; 
but in all cases, when the component changes directed by 
/'and Q are fully effected, the resultant is the same, viz. ac. 
In the problems which follow, we are seldom concerned 
with the actual paths. 

Let A be the initial position of the object in the room ; 
draw AC parallel and equal to ac, and similarly directed, 
then C is the final position of the object. 

The law of the triangle is of fundamental importance ; 
two vector quantities of any like kind are compounded by 
the same rule. The student should pay great attention 
to the rule. Observe that, for compounding P and Q, two 
triangles, which are similar, may be drawn ; that the two 



XXI 



(IRAPHIC STATICS 



507 




p's are similarly directed in the triangle, and also the two 
^'s ; that these directions are with the motion of a point 
which travels round either triangle against the resultant ac. 
This last property may be expressed by saying that in either 
triangle the components p and q are circuital, and the 
resultant r is non-circuital (with/ and q). 

Note also that in the parallelogram abed the three 
vectors p, q, r which meet at a are all directed away from 
a, and those that meet at c are all directed towards c. 

The resultant of P and Q might therefore be found as 
the diagonal r of a parallelogram of which p and q are 
adjacent sides ; />, q and r being directed all towards, or 
all away from, the common point where the three meet. 
This construction is known as the parallelogram of displace- 
ments ; it is exactly equivalent to the triangle of displace- 
ments. 

We thus have the following equivalent rules for obtaining 
the resultant of two given vectors : 

Rule 1.- Place the two given vector lines end to end 
circuitallv, and obtain the non-circuital closing side of the 
vector triangle. 

Rule 2. Obtain the diagonal of a parallelogram which 
//as the two given vector lines as adjacent sides, the three lines 
being all directed away from, or toivards, their common end 
point. 



508 GRAPHICS chap. 

400. The vector polygon. Let P, Q, P, S represent 
any component displacements which a body receives ; it is 
required to find the resultant displacement. 

Starting from any point a, Fig. ( i ), draw ab, be, cd, de 
respectively, equal, parallel, and similarly directed to P, Q, 
P, S; join the first point a to the last point e, and direct 
it from a to e ; then ae represents the resultant. 

For, by the rule of the triangle, ac is the resultant of ab 
and be ; ad is the resultant of ac end cd, and therefore of 
ab, be, cd; and ae is the resultant of ad and de, and there- 
fore of ab, be, cd, de. The vector polygon abede is called 
the polygon of displacements, and ae is its clositig side. 

In drawing the polygon the components may be joined 
in any sequence it will be found that in all the polygons 
which are thus possible the closing sides will be vectorally 
equal (i.e. equal, parallel, and similarly directed). Fig. (2) 
shows the polygon when the sequence is S, Q, P, P; 
the closing side of (2) is seen to give the same resultant as 
the closing side of (1). 

Observe that in both polygons the component sides 
p, q, r, s are directed circuital!) 7 , and the resultant side 
v non-circuitally with the others. 

The polygon (3) has been drawn to show the student that 
if the component sides are not all circuital, the answer is 
incorrect ; the closing side does not then give the resultant. 

The vector polygon may be applied to find the resultant 
of any number of components. 

If the end point e had coincided with the starting-point 
a the resultant would have been zero, and all the sides would 
have been circuital. Thus a body which receives com- 
ponent displacements, represented by the circuital sides of 
any closed polygon, has no resultant displacement. 

The rule for finding the resultant of a given system of 
vectors may be stated as follows : 

Rule. Place the given vector lines end to end circuit- 
ally, and obtain the non-circuital closing side of the vector 
polygon. 



XXI 



GRAPHIC STATICS 



50Q 






Examples. 1. A body receives component displacements of 2, 
5, and 7 units in directions parallel respectively to the sides of 
an equilateral triangle taken circuitally. Find the resultant dis- 
placement. Ans. 4.3. 83.4, 36.6 with 3 and 5. 

2. Draw a quadrilateral ABCD having given AB = 3.J, BC2..'], 

CD =5.2, 75^ = 3.4, AC=^.i t . Suppose a body to be moved 
4 feet parallel respectively to each of the directions AB, BC, 
CD, AD, AC ; find the resultant movement. Show, by draw- 
ing a figure, that if the displacements had taken place in any 
other sequence, say the sequence BC, AD, AB, AC, and CD, 
the resultant displacement would have been the same both in 
magnitude and direction. Ans. 11.4 feet, 6o with AB. 

3. A person walks 29 steps east, 51 steps south-south-east, and 13 

steps south-west. Find how many steps respectively west and 
north will bring him back to his original position. 
Ans. 39.3; 56.3. 

4. A body is moved 10 feet parallel to a line OX, and 15, 20, and 

25 feet each in directions making respectively to 30, 90 , and 
'35 with OX. Find the component displacements parallel and 
perpendicular to OX, and the resultant displacement. 
Ans. 5.31 feet, 45.2 feet, 45.5 feet, 83. 3 . 



510 GRAPHICS chap. 

401. Concurrent systems of forces. All vectors have 
magnitude, direction, and sense. Some, like linear veloci- 
ties and couples, have no particular location in space ; they 
are completely represented by a directed segment of a line 
which is free to be moved parallel to itself either laterally 
or longitudinally or in both ways. Others, such as rotations 
and forces, take place along definite lines ; they are repre- 
sented by a directed segment of the line ; the segment may 
have any position in the line, but may not be moved 
laterally out of the line. 

In the problems which follow, the vectors will be con- 
fined to forces. To specify a force we must define its 
magnitude, direction, sense, and any one point in its line of 
action. 

When there are several forces acting on a body, we 
speak of them collectively as a system of forces. If the 
lines of action of all the forces pass through a common- 
point, we have a concurrent system. In the first instance 
we shall direct attention to such systems. 

The general propositions on unlocalised vectors already 
given apply to forces, with the restriction that forces are 
vectors localised in lines. 

Thus to compound a system of concurrent forces, we first 
construct a' polygon of forces, that is a vector polygon drawn 
as if the forces were unlocalised. The resultant is then 
represented by a line drawn through the common point of 
the system, and equal, parallel, and similarly directed to 
the non-circuital closing side of the force polygon. If the 
force polygon close, i.e. if the first and last points coincide, 
the resultant is zero. In this case the system is in 
equilibrium, the components balancing one another. 

A force which being added to a system of forces pro- 
duces equilibrium is called the equilibrant of the system. 
It is equal in magnitude to the resultant, and acts in the 
opposite direction along the same line. 

We now state some theorems relating to concurrent 
forces. 



xxi GRAPHIC STATICS 511 

Theorem 1. The resultant of a concurrent system of 
forces is represented by a line through the common point, 
equal, parallel, and similarly directed to the non-circuital 
closing side of a force polygon. 

Theorem 2. // is necessary and sufficient for the equili- 
brium of a concurrent system of forces that a force polygon 
shall close. 

Theorem 3. Two forces which are equal in magnitude 
and which act in opposite directions along the same line 
balance one another. 

Theorem 4. Two forces which are in equilibrium must 
be equal in magnitude, and must act in opposite directions 
along the same line. 

Theorem 5. The resultant of two forces passes through 
their intersection, and its magnitude, direction, and sense may 
be obtained by constructing the triangle or the parallelogram 
of forces as described in Rules 1 and 2, Art. 399. 

Theorem 6. Three concurrent forces will be in equilibrium 
if their magnitudes and senses can be represented by the 
circuital sides of a triangle of forces. 

Theorem 7. /// order that three forces mar be in equili- 
brium they must be concurrent {unless they are parallel), and 
their magnitudes and senses must be represented by the 
circuital sides of the triangle of forces. 

Theorem 8. If forces which balance one another be added 
to or subtracted from any system of forces, the resultant of the 
system is unaltered. 

Theorem 9. If a system of forces is in equilibrium, any 
force reversed in sense is the resultant of the others. 

Theorem 10. The algebraical sum of the components of a 
system of forces in any direction is equal to the component of 
the resultant in the same direction. 

These theorems are collected for easy reference, but the 
beginner will only gradually come to understand them as he 
applies them to problems. He should consult them from 
time to time as he works the succeeding examples of this 
chapter. 



512 GRAPHICS chap. 

402. Problem. Forces of 5 and 2 units act respectively 
to the right and upwards along the given lines OM, ON 
which include an angle of 70 . Required the resultant. 

First Method. By the parallelogram of forces. 

Let G be the intersection of the given lines. 

To any convenient scale mark off from O, to the right 
along OM, and upwards along ON, the lengths OP, OQ 
of 5 and 2 units respectively. 

Complete the parallelogram by drawing PR, QR parallel 
to OQ, OP. Draw the diagonal OR, and direct it from 
O to R. 

Then OR represents the required resultant. It measures 
6 units and makes 18.5" with OM and 51. 5 with ON. 

Second Method. By the triangle of forces. 

Draw any line ab parallel to OM, directed to the right 
and 5 units long. From h draw be parallel to ON, directed 
upwards, and 2 units long. Join ac, and direct it from a to 
c, i.e. non-circuitally. 

From O draw OR vectorally equal to ac, that is parallel, 
equal, and similarly directed. Then OR represents the 
resultant as before. 

403. Problem. A given force is represented by OR. 
Required its components along the given lines OM and ON. 

Draw RQ and RP parallel to MO and NO. Direct 
OP and OQ away from O, like OR is directed. 

Then OP and OQ represent the required components. 

404. Problem. A given force of 34 lbs. acts along OG. 
Required its component parallel to LL, which makes 35 
with OG. 

Reciangtdar component is understood. 

Select a convenient scale, say \" to 10 lbs., and mark 
off OG 34 units long. Direct it from O to G. 

Draw Off, GK parallel to LL, and OK, GLL perpen- 
dicular to LL. Direct Off and OR like OG, away from O. 

Then Off represents the required component parallel to 
LL, and measures 27.9 lbs. The perpendicular component 
O K measures 19.5 lbs. 



XXI 



GRAPHIC STATICS 



513 




405. Problem. The rectangular components of a force 
are 7 and 5 lbs. Find the force. 

Draw any two perpendicular lines, along which, to a 
convenient scale, mark off from their intersection O, OA 
and OB to represent 7 and 5 lbs. 

Complete the rectangle contained by OA, OB, and draw 
the diagonal OC. Then OC represents the required force ; 
it measures 8.60 lbs. and makes 35.5 with OA and 54.5 
with OB. 

406. Problem. A small ring is at rest under pulls of 
2, 3, and 4 lbs. exerted through strings. Find the angles 
between the strings. 

Draw a triangle of forces def with, its sides 2, 3, and 4 
units long and circuitally directed. 

From the centre of the ring draw lines vectorally equal to 
the sides of the triangle. 

These represent the three pulls, and the angles D, E, F 
between them are found to measure 133.4 , 7 5. 6, and 15 i. 

2 L 



514 GRAPHICS chap. 

407. Problem. A load of 2 tons is suspended from 
a pin A, which is maintained in position by the pull of a 
horizontal tie AL, and the thrust of a strut MA inclined 
at 25. Find the pull and thrust. 

Draw ab 2 units long and directed vertically downwards 
to represent the given load. From the ends a and b draw 
two lines parallel to the tie and strut to intersect in c, and 
direct the sides be, ca circuitally with ab. 

Then abc is the triangle of forces representing the equili- 
brium of the pin A. The pull of the tie is found by 
measuring ca to be 4.28 tons. And the thrust of the strut 
AT A is 4.72 tons, as given by be. 

408. Problem. Four given pulls and thrusts K, L, M, 
N act on a point as denned in the figure. Find their 
equilibrant. 

Construct a vector polygon thus : Select a convenient 
scale (say |" to 10 units), and from any point a draw ab, 
be, cd, de respectively parallel and similarly directed to K, 
L, M,N, and 72, 40, 105, and 58 units long. 

Join ea and direct it circuitally, i.e. in from e to a. A 
line R drawn through the common point vectorally equal to 
ea represents the required balancing force. 

Measuring ea, the magnitude of R is found to be 88.5 
units. The angles E and A measure 45 and 121. 

Observe the two systems of notation. In one, the forces are 
denoted by A*, L, M, N, R, and the corresponding sides of the force 
polygon by k, I, m, n, r. 

In the other, known as Bow's notation, the angles round the 
point are lettered A, B, C, D, E, and the corresponding corners of the 
force polygon a, b, c, d, e. In this system a force is referred to by 
naming the two letters in the angular spaces on each side of it. Thus 
the force " BC" (or CB) would mean the force "Z," and the corre- 
sponding letters b, c would appear at the ends of the side /. 

On going round the point (clockwise), the sequence .-/, B, C, D, E 
of the angles is seen to be the same as the circuital sequence a, b, c, d, 
e of the corners of the vector polygon. If we agree that the two letters 
which denote any force shall be named in the sequence derived from 
the rotation, e.g. BC for L, then the same sequence, be, agrees with the 
circuital arrow. This convention will be used later. 



XXI 



GRAPHIC STATICS 



515 




Tons 
L\40^76--M=I05 




K=7Z a \ J7 

^ V N-58 



R> 




\ 



Examples. 1. Find the force whose horizontal and vertical com- 
ponents are 72.2 lbs. and 45.6 lbs. Ans. S5.3 lbs. inclined at 
32. 3- 

2. A force of 100 lbs. acts at an angle of 53. 7 with the horizontal : 

find its horizontal and vertical components. Ans. 59.2 lbs ; 
80.6 lbs. 

3. Two forces of 3.5 and 6.7 lbs. act in directions which include an 

angle of 75; find their resultant. Ans. 8.32 lbs., making 24 
with the larger force. 

4. The lines of action of two forces include an angle of 120 ; one 

of the forces is 10 lbs., and their resultant is 9 lbs. Find the 
other force. Ans. 7.4 or 2.62 lbs. 

5. A horizontal string 8 feet long is 

and from a point 3 feet from one 
end a load is hung which is in- 
creased until at 10 lbs. the string 
breaks, the deflection when this 
happens being 6". If a piece of 
the same string were to hang 
vertically, what load would it 
carry? Ans. 37.9 lbs. 

6. Determine the resultant of the con- 

current system of forces defined by 
the figure. Ans. A pull of 54.4 



\7Z 
f/43\ 

128 \* 

making 20. 4' with the 41 force, and 17. 6 with the 63 force. 




two walls, 



9Z 



34 



516 GRAPHICS chap. 

409. Problem. Forces which act at a point are in 
equilibrium ; all except two, M and N, are known com- 
pletely, but only the lines of action of these latter are 
given. To find the magnitudes and senses of M and N. 

The problem is solved by applying Theorem 2, Art. 401. 

We shall again illustrate Bow's notation, and to be able to apply 
this the figure has been first prepared by drawing the lines of action 
of the forces each with one end at the common point, so that there 
shall be the same number of angular spaces as there are forces. It has 
also been arranged that M and N shall be adjacent so as to bound one 
of the spaces ; this is always possible, since a force may be represented 
either as a pull from one side or as an equal push from the other. 

The spaces are then labelled A, B, C, D, X. Clockwise rotation 
being adopted, the construction may be described as follows : 

Draw ab, be, cd the three sides of the force polygon cor- 
responding to the three given forces AB, BC, CD. 

Close the polygon by drawing the two remaining sides 
dx, ax parallel to DX, AX, intersecting in x. 

The required forces M and JVare then given by the sides 
dx, xa taken circuitally, or as determined by the clockwise 
sequence DX, XA. M is seen to be a pull, and N a thrust. 

410. Problem. A load W is suspended from a jib 
crane ; to find the thrust in the jib CA, and the pull in 
the tie AB. 

First consider the equilibrium of the tie. At the end A it is in con- 
tact with the pin, and between the two surfaces there is a force action 
consisting of equal and opposite forces, one on the pin, the other on 
the bar. If the joint is supposed frictionless, the line of action of these 
forces must pass through the centre of the pin. 

A similar force occurs at the other end B of the tie. Neglecting 
its weight, these are the only two forces acting on the bar, since it is 
nowhere else in contact with anything. And as the tw<o forces balance 
they must be equal and opposite. The bar must therefore be subject 
to a direct pull or a direct thrust. The tie is shown separately under 
forces T, T indicating a pull. 

So for the jib. And generally, in a hinged frame, loaded at the 
Joints only, the forces at the ends of any bar must be equal and opposite, 
and must act along the line joining the centres of the pins. 

To obtain these forces, and to distinguish pulls from thrusts, we 
must draw the closed polygons of forces for the pins which hold the 
frame together at the joints. 



XXI 



GRAPHIC STATICS 



5i7 




409 






Consider the forces on the pin A. There is the vertical force JV, 
being the load suspended from it. There is the force which the tie 
exerts on it acting in the line of the tie. And there is the force exerted 
by the jib along the line of the jib. The pin is shown separately with 
the lines of action of these three forces, the angular spaces round the 
pin being labelled A', Y, Z. 

To draw the triangle of forces for tlie pin A. Adopting clockwise 
rotation, the sequence of letters for W, the known force, is X Y. So 
set off xy vertically downwards to represent //". Draw xs, yz parallel 
to XZ, YZ. By watch - hand rotation about the pin obtain the 
sequences YZ, ZX for jib and tie, and direct those lines near the pin 
A to agree with the sequences yz, zx of the triangle of forces. 

YZ or .S" is directed towards the pin A, indicating a thrust in the 
jib ; and ZX or T is directed away from the pin, denoting a///// in the 
tie. The magnitudes are obtained by measuring yz and zx. 



518 GRAPHICS chap. 

411. Problem. A hinged triangular frame is in equili- 
brium under given forces P, Q, R applied at its corners. 
To find the forces in the three bars of the frame. 

Since the forces P, Q, A' maintain equilibrium, their lines must all 
pass through one point (not shown), and their magnitudes and senses 
must be given by the circuital sides of the triangle of forces abc. 

To employ Bow's notation, the lines of the applied forces P, Q, R 
are all drawn outside the frame, and with their ends at the joints. The 
external and internal spaces are then lettered A, B, C, D. 

To obtain the forces in the bars, we may draw the triangles of forces 
for the three joint pins. Clockwise rotation is adopted. 

The tipper joint. The spaces round this joint named in watch-hand 
sequence are A, B, D. The triangle of forces named circuitally is 
abd. The forces BD, DA, directed in the senses bd, da, both act to- 
wards the joint, and indicate thrusts in the two inclined bars. 

The left-hand joint. The spaces in clockwise sequence are C, A, 
D. The triangle of forces is cad. The forces AD, DC, directed as 
ad, dc, act the first towards, the second away from the joint. The 
horizontal bar of the frame is thus in tension. 

The right-hand joint. The spaces are B, C, D, and the triangle of 
forces is bed. 

The force diagram. The four triangles of forces may be superposed 
on one another so as to form the figure abed. The six lines of the 
latter measured to scale give the magnitudes of P, Q, R and of the 
forces in the bars. It is called the force diagram for the frame. The 
arrow-heads would conflict and are omitted, but the sense of any force 
may be obtained by the convention of Bow's notation, rotation about 
the joints being always clockwise. 

412. Problem. A string has its ends fixed, and four 
forces P, Q, R, S act on it along the given lines. The 
form taken by the string is shown in the figure. If P is 
3 lbs., find the other forces Q, R, S, and the tensions in the 
five segments of the string. 

Bow's notation can be applied. The spaces are lettered A, B, C, 
D, E, O. We shall again adopt clockwise rotation. 

Draw ab 3 units long to represent P or AB. And draw ao, bo par- 
allel to AO, BO to intersect in 0. Then abo is the triangle of forces 
for the joint ABO. 

Draw be, oc parallel to BC, OC, to intersect in c. 

Draw cd, od parallel to CD,OD ; and de, oe parallel to DE, OE. 

Then oabede is the force diagram. The forces P, Q, P, S can be 
found by measuring ab, be, cd, de to a scale of -J" to 1 lb. ; and the ten- 



XXI 



GRAPHIC STATICS 



519 



a- d 




Q 412 



sions in the segments A, B, C, I), E of the string by measuring oa, ol>, 
oc, od, oe. 

Definition. The line of the string is called a funicular polygon or 
link polygon. The point is its pole. The several lengths of the 
string may be supposed to be replaced by weightless rigid links, con- 
nected by hinged joints. 

An indefinite number of such link polygons having for their pole, 
or having new poles, could be found, all in equilibrium under the action 
of the same forces P, Q, A\ S ; in some of these the links might be in 
compression. 

The link polygon plays an important part in graphic statics, as the 
succeeding problem will show. 



520 GRAPHICS chap. 



413. Moments of forces. Couples. The constructions 
hitherto Riven are not sufficient in themselves to solve com- 
pletely the general problem on non-concurrent forces in one 
plane. It may be shown that the magnitude, direction, and 
sense of the resultant of such a system is given, exactly as 
before, by the non-circuital closing side of the vector force 
polygon ; but we still require to find the line in which the 
resultant is located : this we can do by obtaining in addition 
to the above any one point in its line of action. 

Forces produce, or tend to produce, rotations in bodies 
as well as motions of simple translation, and we now 
require to measure such a tendency. This is done by find- 
ing the moment as defined in the next paragraph. 

Definition i. The moment of a force about any point is 
the product of the magnitude of the force and the perpendicu- 
lar distance from the point to the line of the force. 

Thus in Fig. (<?) the moment of P about A'Px KM, where P\% 
to be measured on the force scale, and KM on the linear scale. 

It is seen that P tends to cause watctt-hand rotation about K. This 
we may agree to consider as positive ; the opposite tendency is then 
negative. 

Definition 2. A pair of forces equal in magnitude, opposed 
in sense, and acting in parallel lines, is called a couple. 

Definition 3. The moment of a couple is the moment of 
either force about any point in the line of the other. 

Thus in Fig. (/') the forces P, P constitute a couple. The moment 
is Pxp and is positive because the tendency is to rotate clockwise. 

Theorem. Pn any system of fixes the algebraical sum of 
the moments of the components about any point is equal to the 
moment of the resultant about the point. 

This important theorem is proved in books on mechanics. The re- 
sultant moment may be called the moment of the system. 

For example, two downward vertical forces, 8 ft. apart, each of 10 
lbs., have moments of 60 and 20 ft. -lbs. about an intermediate point 
K, distant 6 ft. and 2 ft. from their lines. The algebraical sum is 
60 - 20 = 40 ft. -lbs. Now the resultant is 20 lbs. acting 2 ft. from K, 
and its moment is 20 x 2 = 40 ft. -lbs., the same as before. 



XXI 



GRAPHIC STATICS 



52i 




5ccvt. 



Examples on Problems 409 to 413. 

*1. Copy the jointed frame double size with the supports placed at 
the same level. The loads at the two upper joints, and the 
supporting forces P and Q, are all vertical. Draw the force 
diagram for the frame to a scale of J" to 100 lbs., and measure 
/'and Q. Ans. 235 lbs., 405 lbs. 

*2. Copy the braced cantilever double size. Then draw the force 
diagram to a scale of j" to i cwt. , and measure the forces in the 
four bars of the frame, distinguishing pulls from thrusts. 
Ans. Pulls : 8.37, 20.1 ; thrusts: 6.68, 14. 2 cwt. 

3. Draw a semicircle of 2J" radius with the four equal chords. 
These chords and the diameter form a link polygon. Five 
forces perpendicular to the diameter act at the joints and main- 
tain equilibrium, causing a thrust of 10 lbs. in the long link. 
Draw the force diagram, and determine the forces at the joints 
and the tensions in the four short links. 

4. In Figs. (<r) and (/>) above suppose the linear scale to be J" to 
10", and the force scale " to 100 lbs., measure the distances 
/CI/ and p. and the forces P, P, and calculate the moment of 
P about K in (/7), and the moment of the couple in (/>). 

Ans. 4S00 lb. -ft; 3880 lb. -ft. 



522 GRAPHICS chap. 

414. Problem. To find the moment about K of the 
given force P of 7 lbs., the linear scale for the figure being 
I" to 10". 

Adopting Bow's notation, let the force be known as AB. 

Selecting a suitable force scale (" to i lb. in the figure) 
draw ab 7 units long to represent the given force AB. 

Select any point as pole, at a distance os from ab re- 
presenting unit force, or a force expressed by a simple 
number. In the figure os measures 5 lbs. on the force 
scale. 

Join oa, ob. Through any point p on AB draw the two 
lines, or links, parallel to oa, ob and of indefinite length. 
Those links may be known as links A and B. 

Through A' draw the line parallel to ab, to intersect the 
two links A and B in A , B . 

Then A B x os gives the moment of AB about K, these 
lines being measured by the two scales, one as a length, the 
other as a force. 

^0^0 ao 

Proof. Since the triangles pAJ3 M oab are similar, = ; or 

AqBq x os = ab x pm = P x KM= moment of P about K. 

Moment scale. A moment scale may be found on which 
A Q B can be directly measured. Thus if A Q B were \", it 
would measure 10" on the linear scale, and would represent 
a moment of 10" x os lbs., that is 10" x 5 lbs. or 50 inch- 
lbs. The moment scale is therefore \" to 10" x os lbs., 
that is \" to 50 inch-lbs. ; or more conveniently, V to 100 
inch-lbs. 

Measuring A Q B on this scale, the required moment is 
found to be no inch-lbs. 

Note 1. If K were any other point, the new intercept A B t) , 
measured on the moment scale, would give the moment of P about the 
new A". The figure is thus a diagram of moments. 

Note 2. The diagram is a link polygon for the force /"with respect 
to the pole o. See Definition, Prob. 412. 

If the links were secured each at a point, and hinged together at p, 
they would form two bars of a frame, for which oab would be the force 
diagram. The forces in the bars due to the load P at the joint p 
would be found by measuring oa, ob on the force scale. 



XXI 



GRAPHIC STATICS 



523 




Linear /Scale 

Force 

Moment 

i.e 
or 




414 



< to 10" 

1 4" to /lb. 
M-" to io"*oslbs. 

1 A" to I0"x5 2bs. 

%. " to wo i?ich-lbs. 




415. Problem. To find the moment about K of the 
given couple 8 lbs., 8 lbs., the linear scale being given. 

The linear and force scales of Prob. 414 are used. 

Draw the two sides de, ef of the force polygon, to repre- 
sent the given forces Q, Q, or DE, EF after Bow. 

Take the pole o so that os is V, representing 4 lbs., thus 
giving a moment scale of T V" to 10 inch-lbs. 

Join oe, od, of Draw any link E parallel to oe, and 
where this link meets the lines DE, EF, draw the two links 
D, F of indefinite length, parallel to od, of. 

Draw through K the line parallel to de to cut the two 
links D and F in D , F . 

Then D Q F gives the moment of the couple about K. 
Measured on the scale of T V' to 10 inch-lbs., the moment 
is seen to be 117 inch-lbs. 

Note 1. As before, the link polygon is a diagram of moments. 
The links D and /'are parallel, and the intercept -D ( f is of constant 
length for all positions of A', illustrating the well-known theorem that 
a couple exerts the same turning moment about any point. 

Note 2. Owing to the limited space the figures in the book are 
necessarily small. But the student, with a sheet of drawing-paper at 
his disposal, should, where possible, select scales which give diagrams of 
ample size. 



524 GRAPHICS chap. 

416. Problem. - - To find the resultant of a given non- 
concurrent system of forces in one plane. 

Let P, Q, M, IV ox AB, BC, CD, >E be the forces. 

First, to find the magnitude, direction, and sense of the re- 
sultant. Draw the four circuital sides ah, be, cd, de of the 
force polygon to correspond with the given forces. Join ae 
and direct it non-circuitally. Then the vector ae, when 
localised, will represent the resultant. 

Next, to find a point in the line of the resultant. Choose 
any pole o, and join oa, oh, oc, od, oe. 

Start from any point / on AB. Through p draw the 
link/;- or A of indefinite length 'parallel to oa; draw also 
the link pq or B between AB, BC, and parallel to ob. 
Draw the link qm or C, between BC, CD, and parallel to 
oc, Draw the link inn or D, between CD, DE, and parallel 
to od. And draw the closing link nr or E parallel to oe to 
meet the first link A in r. 

Then r is the required point. A line through r, vector- 
ally equal to ae, represents the resultant completely. 

Reason. If the link polygon were a hinged frame with the forces P, 
Q, M, N, and R reversed in sense, acting at the joints, it will be found 
on examination that oabedeo would be its force diagram, and that each 
of the joints, and therefore the whole frame, would be in equilibrium. 

Note I. Rule for drawing the links. For a balanced system of 
forces, or for a system to which the resultant has been added, observe 
that in Bow's modified notation there are the same number of letters as 
forces, and that each letter is associated with two, and only two, forces. 
And in any complete or closed link polygon the links A, B, . . . are 
parallel to oa, ob, . . . and have their ends on the pairs of forces 
which have A, B . . . common. 

For example, the link B is parallel to oh and terminated by AB, BC 
which have B common. Attention to this rule will prevent mistakes. 

417. Properties of the link polygon. Conditions of 
equilibrium of coplanar forces. 

(a) Partial resultant. The resultant of any portion of the system 
for which the vectors are consecutive in the force polygon can readily 
be found. 

Thus the resultant of BC, CD is represented by a vector equal to 
bd, localised through the point t where the links B, D intersect. 

Observe that Z?,Z> are the first and last letters in the sequence B C, CD. 



XXI 



CRAI'IIIC STATICS 



525 



416 




N /E 



(/>) Moment of the system about any point A". Througli K draw 
a line parallel to ae to meet the links A, E in A w E w Draw os per- 
pendicular to ae. 

Then the required moment = A Q E x os, the scales being determined 
as in Prob. 414. 

(c) Partial moment. The moment of any part of the system for 
which the vectors are added in sequence as in (a) is easily found. 

Thus to obtain the moment of BC, CD about A", draw a line througli 
A" parallel to bd, and draw os' perpendicular to bd. 
Then the required moment = B Q D x os'. 

(d) Parallel forces. If the forces are all parallel the vector polygon 
becomes a line ; the lengths of os, os become equal ; and the lines 
through A", A"' are parallel to the system. 

Thus for parallel forces the link polygon is a very useful diagram of 
moments, the scale being the same for all the intercepts. 

(e) Conditions of equilibrium of a system of forces in one plane. 
The necessary and sufficient conditions are two, viz. 

1. The force polygon must close. 

2. The link polygon must close. 

The first ensures that there shall be no resultant force, though there 
might be a couple. 

The second ensures that there shall be no couple, since when this 
condition holds, the moment of the system about any point is zero. 

By the closing of the link polygon is meant that the last link shall 
intersect the first link on the first force. 

We shall now give some simple problems illustrating the application 
of the above general principles to special cases. 



526 GRAPHICS chap. 

418. Problem. Three parallel forces L, M, N of given 
magnitudes act as shown ; (a) find the parallel forces X 
and Y along the given lines which will balance them ; (b) 
find the resultant of L, M, N ; (c) find the moment of 
the given forces about K ; (d) find the moment of Y and 
L about K. 

The scales are given in the figure, the moment scale being derived 
as explained in Prob. 414. 

Employing Bow's modified notation, the letters AB, BC, CD, DE, 
EA, forming a cycle, are appended to the five balanced forces 
LMNXY. 

{a) To find X and Y. Draw the three sides ab, be, cd of the force 
polygon to correspond with the known forces AB, BC, CD. 

Select a pole 0, and join oa, ob, oc, od. 

Draw the links yl, Im, mn, nx (in Bow's notation named A, B, 
C, D) parallel to oa, ob, oc, od. 

Draw xy {i.e. the dosing link E). And draw oe parallel to xy, thus 
determining the closing sides de, ea of the force polygon. 

The forces X and Y, that is DE, EA, are now found by measuring 
de, ea on the force scale. They are 8 lbs. and 38 lbs., in the senses 
given by the sequences de, ea, that is they both act upwards. 

[b) To find the resultant of AB, BC, CD. The first and last letters 
of this sequence are A and D. 

The resultant acts through r where the links A and D intersect. 
And it is found by measuring ad to be a dozunward force of 46 lbs. 

(r) To find the moment of A B, BC, CD about K. Again observe 
that the first and last letters of this sequence are A and D. 

Draw through K a line parallel to ad, to meet the links A and D 
in A , D . 

Then A D represents the required moment, which measured on the 
moment scale gives the value 61 ft. -lbs. 

Note 1. os was taken 50 lbs. on the force scale, which in conjunction 
with the linear scale leads to the easy moment scale of 1" to 100 ft.- 
lhs. See Prob. 414. 

(d) To find the moment of EA, AB about K. The first and last 
letters are E and B. 

Draw through K a line parallel to eb, to cut the links E and B in 
E B . 

Measuring E^B^ on the moment scale, the required answer is 47 ft.- 
lbs. 

Note 2. The student of applied mechanics will recognise in Fig. 
418 a diagram of bending moments for a beam supported at the ends 
and loaded at intermediate points. See also Fig. 420. 



XXI 



GRAPHIC STATICS 



527 



L=31 M=60 N=45 

A \B K b\c c\d 




Linear scale h'bol' Force scale 3 /i6 'to to lbs. 
Moment scale '/z "to J \ 50 lbs. ? i.e. 1 "to WO ft. lbs 

L g M N 

h 

A\D 





AB 



419 




419. Problem. HKLM is a square of 2h" side. Two 
forces of 9 units each act from H to K, and from L to M, 
forming a couple. A third force of 4 units acts from L to 
K. Find their resultant. 

The scales for the figure are j" to i" and " to I unit of force. 

The given forces are lettered AB, BC, CD. The required resultant 
will be AD. 

Draw the three circuital sides a&, be, cd of the force polygon, and 
the non-circuital closing side ad. 

The pole o is chosen at the intersection of ab and cd. 

The link polygon is begun at A' on AB. The first link A is a line 
through K parallel to oa. The second link B is the point K. The 
third link C is the line AW parallel to oc. The closing link D is a 
line drawn through N parallel to od to meet the first link, which it does 
in A 7 "; so the link D reduces to the point A T . 

Thus the required resultant AD passes through A\ a point on LM 
produced, distant by measurement 3.12" from M, and is a downward 
force of 4 units as given by ad. 



52 8 GRAPHICS chap. 



420. A uniform horizontal rod HK 6 ft. long and weigh- 
ing l. 1 , lbs. is hinged at H, and a downward vertical force 
of 3i"lbs. is applied 2 feet from the hinge. Where must 
an upward vertical force of 2 lbs. be applied to maintain 
equilibrium ; and what will be the pressure on the hinge ? 

The scales for the figure are \" to i' and \" to I lb. 

The weight of the rod may be supposed to act at its middle point. 

Let AB, BC, CD, DA denote the four forces as in the figure, of 
which the first two are given completely, and we require to find the 
line of the third and the magnitude of the fourth. 

Draw the three circuital sides ab, be, cd of the force polygon, to re- 
present the three known magnitudes. The non-circuital closing side 
ad gives a downward force of 3 lbs. for the pressure on the hinge, this 
being the resultant of the other three forces. 

Take any pole o and join oa, ob, oc, od. 

Draw the links A, B parallel to oa, ob, and the two closing links C, 
D parallel to oc, od to meet at t. 

The vertical line tL through t is the required line of action of the 
supporting force, and HL measures 5.75 feet. 

Examples on Problems 414 to 420. 

*1. Copy the diagram double size, and let the linear scale then be 
1" to the foot. Take for the force scale J" to 10 lbs., and let 
the polar distance os be 2", representing 40 lbs. 

(a) Find the resultant of M, P, Q, N. Aus. 18 lbs. upwards 
6.6 feet from Y. 

(b) Find the moment of Af about A". Ans. - 54 lb.-ft. 
(<) Find the moment of the couple P, Q about A'. 

Ans. 80.6 lb.-ft. 
{d) Find the moment of M, P, Q, N about K. 

Ans. -82. 8 lb.-ft. 
((.') Find the forces X, Y which balance M, P, Q, A\ 
Ans. 36.9 lbs. down ; 18.9 lbs. up. 
*2. The jib HJ of a 3-ton crane is inclined at 57 to the horizontal, 
and the tie rod ///at an angle of 27 . Find the thrust in the 
jib and the pull in the tie. Ans. 5.35 tons, 3.27 tons. 

If a back stay IK be added inclined at 45 , and attached to 
the end of a horizontal strut ^/A', find the counter-balance weight 
IV required at K to balance the load on the crane about J. 
Find also the tension in the back stay and the thrust in the 
crane post //. Ans. 2.91 tons; 4.12 tons; 1.43 tons. 

Note. The counter balance weight IV should be found by 
two methods ; first by drawing the force diagram for the frame ; 
next by drawing a link polygon for the three external forces W, 
the load, and the vertical supporting force aty. 



XXI 



GRAPHIC STATICS 



529 



35 1-5 ii?S 

H A\B\C 

? 



a 




5. 



On a horizontal line OX mark off towards X the lengths OA = 
0.5", OB =1.1", 0C= 2.0", 0Z> = 2.5", and above the line set 
off the^ angles OAP= 3S , OBQ=-jo, OCM=no, and ODN 
= 120 . Suppose forces of 320, 145, 570, and 416 lbs. to act 
respectively along PA, QB, MC, and ND. Find the resultant 
force, and the resultant moment about 0, if the linear scale is 
|. Measure and write down (a) the magnitude of the resultant ; 
(/') the angle which its line of action makes with OX ; (c) the 
distance from O (to scale) of the point where the resultant cuts 
OX; (d) the resultant moment of the system about O. 
Ans. (a) 1220 lbs., (b) 94. 3 , (c) 7.T,", (d) 8S60 inch-lbs. 

Draw a rectangle ABCD, making AB= 3 .6", BC=\.^'. 
Forces each of 6.5 lbs. act along AB and CD thus constituting 
a couple. Determine graphically and measure the moment 
of this couple. Ans. 9. 1 inch-lbs. 

In Ex. 4 determine graphically the forces of a couple which, 
acting along the short sides of the rectangle, shall balance the 
given couple. Ans. 2.53 lbs. in the senses AD and CB. 

In Ex. 4 let two additional forces each of 3.7 lbs. act along CB 
and DA. Determine graphically the resultant of the four 
forces, and find the moment of the system about a point A" 
inside the rectangle, distant 0.9" from both AB and CD. 

2 M 



530 GRAPHICS chap. 

421. Centre of gravity. Two examples are now given of 
the determination of the centre of area of a plane figure, 
often termed its centre of gravity. 

The first case represents the cross section of a cast-iron 
beam, symmetrical about a vertical axis YY. The figure 
is divided into three rectangles, the areas of which are 
calculated from the data, supposed given. 

Now through the centres of these rectangles draw parallel 
vectors AB, BC, CD in any direction other than that of 
YY (but at right angles to FKfor convenience), and let the 
magnitudes of the vectors be proportional to the calculated 
areas. Draw the vector polygon abed, and a link polygon 
to any pole o, and thus determine the resultant vector AD, 
passing through the intersection of the closing links. 

The centre of area required is the point G, where the 
resultant vector AD intersects YY, the axis of symmetry. 

In the second example, that of a section of angle iron, 
there being no axis of symmetry, it is necessary to draw two 
link polygons, the parallel vectors, AB, BC, of one having 
a direction differing from those, A'B', B'C, of the other. 
The intersection of the resultants AC, A'C gives G, the 
required centre of area of the section. 

Examples. 1. In the upper figure opposite let the dimensions of 

the top flange be 4" by 1^"; of the bottom flange 9" by i-J"; and 

of the vertical web 10" by 1". Determine the distance of the 

centre of area from the top. Am. 8". 
2. In the lower figure let the angle iron be 4" by z\" outside, and 

the thickness i". Find the centre of area. Aiis. 1.42" and 

0.67" from the top and left sides. 
3- A uniform straight wire 4^" long is bent at right angles at a point 

1 1" from one end ; find its centre of gravity. Ans. j" and 1 

from the long and short sides. 

4. A straight wire 6" long is bent at points distant 1.2" and 3.2 

from end, forming three straight lengths, the first and last being 
parallel, and both making 6o Q with the middle segments. Find 
the centre of gravity of the wire. 

5. A uniform wire 6" long is bent into the form of a semicircle with 

the diameter. Find its centre of gravity. 

6. Find the centre of area of the segment of a circle of 2" radius, 

the chord being 3^" long. 






XXI 



GRAPHIC STATU S 



53i 



T) 


Y 

1 






!-H - - - 


C 


1 

1 
1 

1 




G 

B 


k 


A \ 










532 GRAPHICS chap. 

422. Problem. Known forces are balanced by two 
others, the line of action of one of which, and a point in 
the line of action of the other, are given. To find the 
magnitude and sense of the first balancing force, and the 
magnitude, line of action, and sense of the other. 

Let AB, BC, CD be the known forces ; and let YYbe 
the given line of action of one of the balancing forces, which 
call DE, and P the given point in the line of action of the 
other ; this latter, on Bow's system, must be labelled EA. 
To find the magnitude and sense of DE, and the magni- 
tude, line of action, and sense of EA. 

Draw the force polygon so far as the data will admit, 
viz. the sides ab, be, cd, and an indefinite side through d 
parallel to DE. The solution consists in finding the point 
e in this line. 

Take any pole o, and join oa, ob, oc, od. Draw the link 
polygon with respect to this pole, beginning at the given 
point P. Although the line of action of the force at P is 
unknown, yet P is a point on it, and we may therefore 
begin the link polygon at P. Call y the point where the 
fourth link D meets YY. Join Py in order to close the link 
polygon. Draw oe parallel to Py, and join ea to close the 
force polygon. Then de gives the required magnitude and 
sense of the balancing force DE along YY; and ea gives 
the magnitude, line of action, and sense of EA, the balan- 
cing force which acts through P. 

This problem occurs in connection with roof trusses 
provided with expansion rollers at one end, and subjected 
to the pressure of the wind. 

423. Problem. Any number of known forces are 
balanced by three others which act along given lines. To 
find the magnitudes and senses of the three balancing 
forces or reactions. 

Let AB, BC, CD be the given forces, and XX, YY, 
ZZ the given lines along which the balancing forces act. To 
find the magnitudes and senses of the latter. 



XXI 



CRAI'HIC STATICS 



533 




Denote the required forces by DE, EF, FA as shown. 

Draw the force polygon so far as the data allows ; that is 
draw ab, be, ed, and the indefinite sides through d and a 
parallel to DE and AF The problem is solved when we 
have found e and /on these two lines. 

Take any pole o, and join oa, ob, oe, od. Begin the link 
polygon at one of the intersections of the lines X, Y, Z, say 
at p. The link F then reduces to a point. Draw the links 
A, B, C, D, terminating at x on XX. 

Close the link polygon by drawing px which is the link E. 
Then close the force polygon by drawing oe parallel to px or 
E, and ^/"parallel to EF. 

Then de, ef, fa taken circuitally give the required magni- 
tudes and senses of the balancing forces DE, EF, FA 
acting along XX, YY, ZZ. 



534 



GRAPHICS chap. 



424. Miscellaneous Examples. 

*1. The figure shows a buttress subject to the forces of its own 
weight and a thrust of 2800 lbs. at its upper end. Draw the 
" line of resistance " for the structure. 

Hint, This is a link polygon for the forces, beginning at 
the point of intersection of the forces 2800 lbs. and 400 lbs. 
Compare Prob. 412. 

*2. A heavy uniform wire is bent into the form ABCD, and is sus- 
pended by a string attached to the point A. Draw the direction 
of the string. (1892) 

*3. The directions and magnitudes in lbs. of five unequal forces 
acting at a point a are given. Determine the direction and 
magnitude of their resultant. 0884) 

*4. Five given forces act as shown at a point. Determine by 
construction two forces acting along the given dotted lines which 
will keep the point in equilibrium. Write down the magni- 
tudes and indicate the directions of these forces. Use a scale 
ofi"=ioolbs. (1887) 

*5. Resolve each of the given forces P and Q (P~- 7 lbs., = 9 lbs.) 
along and perpendicular to the given line AB, and write down 
the resultant force in each direction. (1S85) 

*6. A uniform beam AB, weight 44 lbs., is suspended by two equal 
strings from its extremities to the point 0. A weight /' ( = 1 3 
lbs.) is hung on the beam in the given position. Determine the 
position of equilibrium of the system.; ('891) 

*7. A weight of 5 tons is suspended from a, the apex of a triangle 
formed of two bars ba, ca fixed in a vertical wall. Determine 
and write down the stresses in the bars ba, ca. (1893) 

8. The wire passing round the top of a telegraph pole is horizontal, 
and the two directions make an angle of 1 io with one another. 
The pole is supported by a wire stay inclined at 45 to the hori- 
zon. Given the tension of the telegraph wire to be 200 lbs., 
find that of the stay. Ans. 324 lbs. (1896) 

9. ABCD is a square, the angular points being lettered in order. 
Two forces, of 10 units each, act from .-/ to B and from C to 
D, forming a couple. A third force of 15 units acts from C to 
A. Find their resultant. (1892) 

10. Draw six lines oa, ob, oc, od, oe, of, radiating from a point o, 
any two adjacent lines including an angle of 6o Q . These six 
lines are respectively the lines of action of forces of 80, 100, 90, 
60, 120, and 50 lbs. all acting towards except those along oa 
and of, which act away from 0. Determine and write down the 
magnitude and direction of the resultant of the forces. 

Ans. A pull of 1 15 lbs., making 4.3 and 55.7 with oc and od. 



XXI 



GRAPHIC STATICS 



535 



Cvfw //ie /imwvs dwMe size 




536 GRAPHICS 



CHAT. 



*11. Determine the line of action, and write down the magnitude 
of the resultant of the five given parallel forces acting in one 
plane in the directions shown by the arrows. (1888) 

*12. Four vertical forces act downwards as follows : 

At A 10 lbs., at B 18 lbs., at C 16 lbs., and at D 12 lbs. 
Determine the position of their resultant. 

Supposing the two forces at B and C only to act downwards, 
determine the values of two vertical forces acting upwards 
through A and D so as to make equilibrium with those through 
B and C. Employ for the force polygon the scale o. 1" to 2 
lbs. (1895) 

*13. Find (and write down) the moment in foot-tons of the result- 
ant of the pairs of parallel forces, A and B, C and D, with 
regard to the point E. Scale of forces, 0.25" per ton ; scale of 
distances, o. 1" per foot. ( J S94) 

*14. A force of 14! lbs. acts along the given line ab. Determine 
by construction what force acting along cd will have the same 
moment about P. (1886) 

*15. A uniform rod AB, weighing 53 lbs., is pivoted at A. If a 
force P of 32 lbs. is applied at C, where must a parallel force of 
41 lbs. be applied to maintain equilibrium? 

*16. Obtain by construction a line representing the moment of the 
resultant of the two given forces P, Q, about the point 0, using 
a scale of |"= 1 lb., and linear scale full size. (1888) 

*17. / > 1 =i5o lbs., P., = 200 lbs., P 3 =i20 lbs., are three forces 
acting in the direction indicated by the arrow-heads. Find by 
the funicular polygon the resultant moment of the three forces 
acting round the point S. 

Scale of forces, 1 00 lbs. = I inch. Scale of lengths, 50 feet 
= 1 inch. (1897) 

18. A bar of uniform thickness inclined at an angle of 30 to the 
horizontal, with one end against a wall, rests across a rail at a 
point 2 feet away from that end. Find the length of the bar if 
the rail and wall are both smooth. Ans. 5 feet 4 inches. 

Hint. The three forces which act on the bar are ( 1 ) the 
force from the wall, which is in a direction at right angles to 
the wall, (2) the force from the rail which is at right angles to 
the bar, (3) the weight of the bar which acts vertically through 
its middle point. Now use Theorem "] , Art. 401. 

19. Three forces of 11, 19A, and 26 lbs. act at a point Pin such 
directions that their resultant is nil. Draw lines representing 
the forces in direction and magnitude. (1886) 



XXI 



GRAPHIC STATICS 



557 



Cbftv /he fhures dot idle- sue- 













A 


B 





^ 



^ 


% "s 
^ s 





r 




' 


' 


1 






I 



11 



A 

Tons 4 



B 



C D, 



Tons 2 Tons 5lTons 



13 



,E 




P 



P 



15 



C 



C 



D 




B 



S 



17 



?* 




APPENDIX I 

SCIENCE AND ART DEPARTMENT EXAMINATION 

May 1S99 

Advanced Stage 

Instructions 
Only eight questions are to be attempted. 



Plane Geometry. 

*21. C is one vertex of a triangle ; (J is the centre of the circum- 
scribed circle ; the centre of the inscribed circle. Draw the triangle. 

(22) 

22. Draw an indefinite line PC. At any point in it, N, draw a 
line perpendicular to BC, and set off from N on the perpendicular, 
above and below BC, lengths NP, NP 2 , each equal to 2.4 inches. 
BC is the axis of a parabola, and P, P 2 are points on the curve. Calling 
A the vertex of the parabola, draw the curve such that the area 
bounded by the double ordinate PP 2 and the portion of the curve 
/'A P., shall be 4 square inches in area. (20) 

*23. Two equal elliptic wheels, A and B, are in contact at 0. 
Their foci are : those of A, I\, and E, ; those of B, f v and /.,. The 
wheel A is driven by the wheel B ; and the wheels rotate round fixed 
axes at the foci I\ andy^ respectively. A pin is attached to A at its 
focus F 2 . Draw a diagram representing the vertical heights of the pin 
above the line CD during a complete revolution of the wheels, the pin 
at the commencement of the revolution being at the point F 2 on the 
diagram. Take for abscissae \ inch to represent T Vth of a complete 
revolution ; and for ordinates the actual heights of the pin corre- 
sponding, above CD. The position should be shown for, at least, 
every T Vth of a complete revolution. (22) 

Solid Geometry. 

*24. a!>, db' are the projections of a line AB ; I'h, hgzxe. the traces 
of a plane. Draw the projections of a line in the plane, meeting AB, 
and making with it an angle of 35 . (24) 



AI'PEN. I 



EXAMINATION PAPER 



539 



The diagrams (ejccenbN?26)tobeprikedoff or (7899) 
accurate/ f transferred to the paper. 



Q 

G 



c 

O 



A 

/ \ 

/ \ 



C / \ 7) 




540 ADVANCED STAGE appen. 

25. A cube, inscribed in a sphere of i^ 5 inches radius, has two 
adjacent faces inclined at 30 and 70 respectively, to the horizontal 
plane. Draw the solids in plan and elevation. (24) 

*26. Draw an equilateral triangle abc (see diagram, which is not 
drawn to scale) of 4.2 inches side, and set off lengths aA, aB, cF, etc. , 
1 inch distant from each vertex, along the sides. The figure ABCDEF, 
so formed, is a horizontal section of a regular octahedron, lying with 
one face on the horizontal plane. Draw the octahedron, showing on it 
the outline of the section in plan, and its height above the horizontal 
plane in elevation. (26) 

*27. A and B are the scales of slope of two planes. Draw the plan 
of a sphere of I inch radius, resting on the horizontal plane, and touch- 
ing the two planes, both of which pass over the sphere. Show the points 
of contact with the planes, and write down their figured heights. 
Unit = o. 1 inch. (24) 

*28. ad ', bb', cc' are the projections of three points A, B, and C. 
Find the projections of a fourth point D, distant 2j inches from each 
of the points A, B, and C. Complete the tetrahedron formed by 
joining the four points. (22) 

29. The vertex of a cone is 1.5 inches above the horizontal plane, 
and its axis is inclined at 45". Its generating lines make an angle 
of 30 with the axis. Determine the scale of slope of a plane tangent 
to the cone, and inclined at 6o u . Show the line of contact on the cone 
in plan. Unit = 0.1 inch. (24) 

*30. The diagram represents a cone AB V, lying on a block DEFG 
whose thickness is DD 2 . Draw on the plan the outline of shadow 
thrown by the solids, one on the other, and on the horizontal plane 
of projection. Show also on the plan the limit of light and shade on 
the cone. The arrows indicate the direction of the parallel rays of 
light, inclined at 45 to the xy line in plan and elevation. (30) 

*31. Draw the solids of Question 30 in isometric projection. The 
vertical isometric planes to be taken parallel to the planes DE and 
DG, and the vertical through D nearest to the observer. An isometric 
scale must be employed. (24) 

Graphic Statics. 
Alternative and Optional. 

32. A right truncated prism has for base an equilateral triangle of 2 
inches side. The three edges perpendicular to the base are respec- 
tively 2 inches, 1.75 inches, and 1.25 inches in length. Find, by 
graphic construction, a line representing the cubic contents of the 
solid, to a unit of 1 inch. (20) 

*33. ABCD is a horizontal rigid bar, hinged at A, loaded ati? with 
40 lbs., at 6' with 50 lbs., and retained in place by a cord DE (passing 
over a pulley) attached to a pin at D. Find the stress in the cord DE. 
Employ the funicular polygon, using, for the scale of loads, J inch to 
represent 10 lbs. ( 2 4) 



EXAMINATION PAPER 



541 



(1899) 




Note. The figures are reproduced half size. 



APPENDIX II 



DEFINITIONS AND THEOREMS OF PURE SOLID GEOMETRY 

Definitions 

Definition 1. & plane is a surface such that any two points being 
taken in it, the straight line joining them lies wholly in the surface. 

Definition 2. Parallel planes are such as do not meet each other, 
though produced. 

Definition3. A straight line is parallel to a plane when the two 

do not meet each other, though produced. 

Definition 4. A straight line is perpendicular to a plane, when it 
is perpendicular to every straight line which meets it in that plane. 

Definition 5. Two planes are perpendicular to each other when 
any straight line drawn in one, perpendicular to the intersection of the 
planes, is perpendicular to the other plane. 

Definition 6. The orthographic projection of a point on a plane 
is the foot of the perpendicular from the point to the plane. The per- 
pendicular is called a projector, and the plane is called thep/ane of pro- 
jection. 

Definition 7. The orthographic projection of a given line on a 
plane is the line generated by the foot of a perpendicular to the plane, 
which perpendicular moves so as always to intersect the given line. 

The surface generated by the moving perpendicular is called the 
projecting surface. When the line which is projected is straight, the 
projecting surface is called the projecting plane. 

Definition 8. The inclination of a straight line to a plane is the 
angle between the line and its orthographic projection on the plane. 

Definition 9. The angle between two planes, called a dihedral 
angle, is measured by the angle between two straight lines, drawn from 
a point in their intersection, each perpendicular to the intersection, and 
lying one in each plane. 



appex. ii DEFINITIONS 543 

Definition 10. The inclination to each other of two straight lines 
in space which do not intersect is measured by the angle between two 
lines from any point, respectively parallel to those lines. 

Definition 11. A solid ox polyhedral angle is formed when three or 
more planes meet in a point. It consists of as many plane angles, and 
also of as many dihedral angles as there are planes. 

Definition 12. A solid is that which has length, breadth, and 
thickness. It is completely bounded by a surface, or by surfaces, which 
may be plane or carved. 

Definition 13. A. polyhedral is a solid bounded by plane surfaces, 
called the faces, which meet in straight lines called the edges. 

Definition 14. A prism is a polyhedron of which the side faces 
are parallelograms, and the two end faces, or bases, are similar and equal 
polygons in parallel planes. The line joining the centres of the bases 
is called the axis of the prism. 

If the axis be perpendicular to the base, the prism is said to be a 
right prism ; if not, it is said to be oblique. 

The perpendicular distance between the bases is called the altitude. 

Definition 15. A pyramid is a polyhedron, one" face of which, 
called the base, is a polygon, the other faces being triangles which have 
a common vertex. 

The common vertex of the triangles is called the vertex of the 
pyramid, and the line joining the vertex to the centre of the base is 
called the axis of the pyramid. If the axis be perpendicular to the 
base, the pyramid is said to be a right pyramid ; if not, it is said to be 
oblique. The perpendicular distance from the vertex to the base is 
called the altitude. 

Definition 16. A pyramid having a triangular base is called a 
tetrahedron. 

Definition 17. A polyhedron is said to be regular when its faces 
are similar, equal, and regular polygons, and all its dihedral angles are 
equal to one another. 

Definition 18. A regular tetrahedron is a solid having four equal 
and equilateral triangles for its faces. 

Definition 19. A cube is a solid having six equal squares for its 
faces. 

Definition 20. A regular octahedron is a solid having eight equal 
and equilateral triangles for its faces. 

Definition 21. A regular dodecahedron is a solid having twelve 
equal and regular pentagons for its faces. 

Definition 22. A regular icosahedron is a solid having twenty 



544 PURE SOLID GEOMETRY appen. 

equal and equilateral triangles for its faces, and all its dihedral angles 
equal. 

Definition 23. A surface of revolution is the surface generated 
by the rotation of a line (straight or curved), about a fixed straight line, 
to which it is supposed to be rigidly connected. 

The fixed straight line is called the axis, and the rotating line the 
generator. 

Definition 24. A conical surface is that generated by a straight 
line which moves so as to always pass through a fixed point, and to 
intersect a fixed curve in space. 

The fixed point is called the vertex. The fixed curve is called the 
directing curve. 

Definition 25. A conical surface of revolution is the surface 
generated by one of two intersecting straight lines, rotating about the 
other as axis, the lines being supposed rigidly connected together. 

The point of intersection of the lines is called the vertex, and the 
complete surface consists of two portions extending indefinitely, one on 
each side of the vertex. 

Definition 26. A right circular cone is the solid generated by 
the revolution of a right-angled triangle about one of the sides con- 
taining the right angle as axis. The circle generated by the other of 
the sides containing the right angle is called the base of the cone. 

Definition 27. A cylindrical surface is that generated by a straight 
line which moves so as to be always parallel to a fixed straight line, 
and to intersect a fixed curve. 

Definition 28. A right circular cylinder is the solid generated by 
the revolution of a rectangle about one side as axis. 

The two sides of the rectangle which are perpendicular to the axis 
generate circles, each of which is called a base of the cylinder. 

Definition 29. A sphere is the solid generated by the revolution 
of a semicircle about its diameter as axis. The centre of the sphere is 
the point which is equidistant from all points in the surface. 

Theorems 

Theorem 1. The plane which contains two parallel straight lines 
will also contain any third straight line which intersects them. 

Theorem 2. A plane can be found which shall contain two inter- 
secting straight lines. 

Cor. A plane is determined by three intersecting straight lines AB, 
BC, CA not concurrent, or by three points A, B, C not collinear. 

Theorem 3. If two planes cut each other, their intersection is a 
straight line. 



II THEOREMS 545 



Theorem 4. If a straight line be perpendicular to each of two 
other straight lines at their point of ' intersection t it will be perpendicular 

to the plane containing the t-wo lines. 

Theorem 5. If one of two parallel straight lines be perpendicular 

to a plane, the other line will also be perpendicular to the plane. 

Theorem 6. If a straight line be perpendicular to a plane, every 
plane containing the line -will be perpendicular to that plane. 

Theorem 7. The orthographic projection of a finite straight line 
on a plane is the straight line joining the projections of its ends. 

Theorem 8. If three or more straight lines which meet at a point 
are each perpendicular to the same straight line, they are in one plane. 

Theorem 9. If two or more straight lines are perpendicular to the 

same plane, they are parallel to one another. 

Theorem 10. If two or more straight lines are parallel to the same 
straight line, they are parallel to one another. 

Theorem 11. If two straight lines -which meet are respectively 
parallel to two others which meet, but are not in the same plane, the 
first two and the other two will contain equal angles. Also the plane 
containing the first two is parallel to the plane containing the other two. 

Theorem 12. Planes which are perpendicular to the same straight 
line are parallel to one another. 

Theorem 13. If two or more pa7-allel planes be cut by another 

plane, the lines of intersection are parallel to one another. 

Theorem 14. If two or more straight lines be cut by parallel 
planes, they will be cut in the same ratio. 

Theorem 15. If two intersecting planes be each perpendicular to 
a third plane, their intersection will also be perpendicular to that plane. 

Theorem 16. If two planes which meet be cut by a third plane 
-which is perpendicular to their line of intersection, the lines in which 
the third plane cuts the other two will both be perpendicular to the inter- 
section of the other two. 

Theorem 17. If two or more straight lines be parallel to one 
another, their projections on any plane will also be parallel to one another. 

Theorem 18. If a finite straight line be parallel to a plane, the 
lengths of the line and its projection on the plane are equal to each other. 
If the line be inclined to the plane, the length of the projection is less 
than the length of the line. If the line be perpendicular to the plane, 
its projection is a poi>it. 

Theorem 19. If a straight line be divided into two or more seg- 
ments, their projections on any plane (iwt perpendicular to the line) are 
in the same ratio as the segments themselves. 

2 N 



546 PURE SOLID GEOMETRY ArPEN. 

Theorem 20. If two planes intersect, and a straight line perpen- 
dicular to one be projected on the other, the projection is perpendicular 
to the intersection. 

Theorem 21. If two straight lines be perpendicular to each other, 
their projections on any plane parallel to one of them will also be perpen- 
dicular to each other. 

Theorem 22. If the projections of two straight lines on any plane 
be perpendicular to each other, and one of the lines is parallel to the plane 
of projection, the two lines are perpendicular to each other. 

The Sphere and the circular Cone and Cylinder. 

Theorem 23. The projection of a sphere on any plane is a circle 
of diameter equal to the diameter of the sphere, the centre of the circle 
being the projection of the centre of the sphere. 

Theorem 24. Any plane section of a sphere is a circle, the centre 
C of which is the foot of the perpendicular from the centre of the sphere. 

Note. If the plane contain O, the section is called a great circle ; 
other sections are small circles. 

Theorem 25. The intersection of two spheres is a circle whose 
plane is perpendicular to the line joining the centres of the spheres, the 
centre of the circle being in this line. 

Theorem 27- The tangent plane to a sphere, centre O, at any 
point P on its surface is perpendicular to OP. 

Theorem 28. A line tangential to a sphere, centre 0, at any point 
P on its surface is perpendicular to OP. 

Theorem 29. If two spheres touch, the point of contact is in the 
line joining their centres. 

Theorem 30- The section of a cone or cylinder by a plane perpen- 
dicular to the axis is a circle whose centre is in the axis. 

Theorem 31. The intersection of a cone or cylinder by a sphere 
whose centre is in the axis consists of two circles with their planes per- 
pendicular to the axis and their centres in the axis. 

Theorem 32. /// a cone or cylinder a sphere can be inscribed 
which lias any given point on the axis as centre, or which passes 
through any given point P on the surface. The curve of contact is a 
circle whose centre is in the axis and whose plane is perpendicular thereto. 

Theorem 33. A cone of indefinite length may be defined by any 
two inscribed spheres, or by one such sphere and the vertex. 

Theorem 34. A cylinder of indefinite length may be defined by 
any two inscribed spheres, or by one such sphere and the axis. 



n THEOREMS 547 

Theorem 35. The projection of a cone or cylinder of indefinite 
length on any plane consists of two lines which touch the projections of 
any two inscribed spheres. 

Note. For the cylinder, if the plane be perpendicular to the axis, 
the projection is a circle which is an edge view of the surface. For the 
cone, if the projection of one inscribed sphere fall within that of the 
other, the surface has no definite form of projection. For illustrations 
of Theorem 35, see the figure on page 321. 

Theorem 36. The projection of a cone or cylinder {or of any other 
surface of revolution) maybe determined as the envelope of the projections 
of its inscribed spheres. 

Theorem 37- Any plane containing the axis of a cone or cylinder 
cuts the surface in a pair of generating lines. 

Theorem 38. The tangent plane to a cone or cylinder at any point 
P on its surface is perpendicular to the plane which contains P and the 
axis. The tangent plane has line contact with the surface, this line 
being the generator through P. The axial plane through P cuts the sur- 
face in the line of contact. 

Theorem 39. Any tangent plane to a cone or cylinder touches all 
inscribed spheres. 

Theorem 40. Any plane which contains the vertex of a cone and 
which touches an inscribed sphere also touches the cone. 

Theorem 41. Any plane zvhich 'is parallel to the axis of a cylinder 
and which touches an inscribed sphere also touches the cylinder. 

Theorem 42. If two cones, or two cylinders, or a cone and cylinder 
circumscribe the same sphere, the intersection of their surfaces is a pair 
of ellipses whose planes are perpendicular to the axes of the surfaces. 

This is found to be a very useful theorem. 

Theorem 43. A common tangent plane to two cylinders can in 
general be found {a) when their axes are parallel ; (b) when they cir- 
cumscribe the same sphere. 

Note. The tangent plane may be determined as the plane which is 
parallel to the axes, and which in (a) touches any two spheres inscribed 
one in each cylinder ; and in (b) touches the common sphere. 

Theorem 44. A common tangent plane to two cones can in general 
be found (a) when their axes are parallel and their vertical angles equal ; 
(b) when they have a common vertex ; and (c) when they both circum- 
scribe the same sphere. 

Note. The tangent plane may be determined as the plane which in 
(a) contains the two vertices and touches any inscribed sphere; in (b) con- 
tains the common vertex and touches two spheres inscribed one in each 



548 PURE SOLID GEOMETRY appen. 11 



cone ; and which in (<) contains the vertices and touches the common 
sphere. 

Theorem 45. A common tangent plane to a cone and cylinder can 
be found when they circumscribe the same sphere. 

Note i. The tangent plane may be determined as the plane which 
contains the vertex of the cone ; is parallel to the axis of the cyliftder, 
and which touches the common sphere. 

Note 2. In theorems 43, 44, and 45 there may be impossible cases. 
Moreover the conditions stated are not exhaustive. 

Theorem 46. The normal from any point to the surface of a cone 
or cylinder lies in the plane determined by the point and the axis, and 
is perpendicular to one of the generators in which this axial plane cuts 
the surface. 

Theorem 47. If a sphere touch a cone or cylinder, the point of 
contact lies in the generator determined by the axial plane through the 
centre of the sphere. 

Theorem 48. Two cones, two cylinders, or a cone and cylinder 
which touch may have line contact or point contact. 

If they have line contact their axes must intersect or be parallel ; and 
the line of contact is the generator determined by the common axial plane. 

If they have point contact at P, the common tangent plane at P con- 
tains the two generators through P. 

Theorem 49. The shortest path on a sphere between two points is 
the smaller arc of the great circle through them. 



INDEX 



The references are to pages 



Abscissa, 150 

Addition, graphic, 487 

Adjustment of set-squares, 2 

Alignment, true, 2 

Amplitude, 128 

Analytical geometry, 156 

Angle between line and plane, 249 

between traces of a plane, 246 

between two lines, 246, 248 

between two planes, 249, 284 

chord of, 19 

how to measure an, 22 

pitch, 476, 478 

to set off an, 22 
Angles, tables of sines, etc. , 20 

trihedral, 179, 466 

ways of denning, 18 
Appendix I., 538 

II-. 542 

Archimedian spiral, 124 
Arc, length of a circular, 112 
Area, centre of, 530 

mean ordinate of, 44 

of polygon, 36, 492 
Arithmetic, graphic, 486 
Asymptotes, 118, 126 

of hyperbola, 104 
Auxiliary circle, 78 

circle, major, 78 

circle, minor, 78 

elevation, 196, 197 

plan, 196, 197 

plane, model of, 19b 

projections, 196 



Axes, co-ordinate, 150, 178 

isometric, 444 

metric, 443 

trimetric, 443, 454 
Axis, conjugate, 104 

polar, 186 

transverse, 104 

Black thread used in plotting, 170 
Board, drawing-, 3 
Bow's notation, 514 

Cam, heart-shaped, 144 

Cams, 142 

Cartridge paper, 3 

Cast shadows, 406 

Celluloid, 120 
ruled, 10, 12 

Centre of area, 530 
of curvature, 90 
of curvature of cycloid, 116 
of curvature of epicycloid, 116 
of curvature of hypocycloid, 116 
of curvature of involute, 122 
of curvature of roulette, 112 
of gravity, 530 

Chart, price, 167 

Chisel-edged pencil, 4 

Choice of scales, 168 

Chord of angle, 19 

< 'hords, scale of, 22 

Circles and lines in contact, 50, 60-66 

Circuital, 507 
non-, 507 



55 



PRACTICAL GEOMETRY 



Circular arc, length of, 112 
Circular protractor, 3 
Clinograph, 3 

how to use, 5 
Closing of force polygon, 510, 525, 

532. 533 

of link polygon, 525, 532, 533 

side of vector polygon, 508 
Co-latitude, 187 
Compasses, 3 
Complete quadrilateral, 34 
Component motions, 130 
Components, 504 

rectangular, 504 
Compounding vectors, 504 
Concrete number, 486 
Concurrent forces, 510 
Conditions of equilibrium, 525 
Cone and inscribed sphere, 320, 342 
Cone, axis of, 70 

double, 70, 71 

generator of, 70 

of indefinite length, 320 

trace of, 322, 326 

vertex of, 70 
Cones, tangential properties of, 255 
Conical surface, 544 
Conic sections, 70 

classification of, 70 

definition of, 72 

properties of, 72 
Conjugate axis of hyperbola, 104 

diameters of ellipse, 86 
Connecting rod of engine, 138 
Construction of ellipse, 76, 82, 83, 
88, 96-98 

of parabola, 99, 102 

of regular polygons, 52 

of triangles, 53-59 
Contact of lines and circles, 50 

of surfaces, 368 
Continued product of lines, 490 
Contour road-map, 167 
Contours, 442, 472 
Co-ordinates, axes of, 150, 178 

of a point, 148-151, 178 

origin of, 150, 178 

polar, 149, 186 

positive and negative, 151, 178 



Co-ordinates, rectangular, 149, 178 

r, 6, <p, 186 

systems of, 149 
Correction of errors of observation, 

170 
Cosine, definition of, 19 
Cosines, table of, 20 
Couples, 520, 523 
Crane test, 170 

jib, 517 
Cube, 543 
Curvature, 90, 91 

centre of, 90, 91 

circle of, 90, 91 

radius of, 90, 91 
Curve, envelope of, 120 

equation to, 154 

e volute of, 122 

involute of, 122 

logarithmic, 126 

of sines, 128 

parallel, 120, 144 

plotting, 157 
Curved surfaces, 308 
Curves, French, 3 

peculiarities of, 118 

projection of, 308 

special, no 

tortuous, 384 
Cusps, 118 

Cutting, earthwork, 472 
Cycloid, 114 

construction of, 114 

curtate, 114 

normal to, 116 

prolate, 114 

tangent to, 116 
Cycloidal curves, 114 

special cases of, 117 
Cylinder, right circular, 544 

of indefinite length, 320 

trace of, 324, 326 
Cylindrical surface, 544 

Definitions, 542-544 
of similar polygons, 26 
of sine, cosine, tangent, 18 

Degree of accuracy in drawing, 2 

Descriptive geometry, 176 



INDEX 



55i 



Development of cone, 312 

of octahedron, 198 

of pyramid, 216 

of sphere, 470 
Diagonal scales, construction of, 14 
Diagram, force, 518 

of load-elongation, 166 

of moments, 522, 526 
Direct-acting steam engine, 138 
Directed quantities, 502 
Directrix, 72 

Displacement of slide valve, 169 
Displacements, triangle of, 399 
Dividers, 3 
Division, graphic, 487 

of a given line, 10, 12 

of line, internally and externally, 

of polygon, 37 
Dodecahedron, 462 
Double cone, 70, 71 
Draughtsman, 2 
Draughtsmanship, good, 177 
Drawing-board, 3 
Drawing, errors in, 2 

-paper, 3 

to scale, 1 

Earth's equator, 186 
Earthwork, contoured, 472 
Eccentricity, 72, 75, 105 
Edge, straight-, 2 
Efficiency-resistance curve, 171 
Effort-resistance curve, 171 
Elevation, front and side, 182 
Ellipse, 70, 74 

auxiliary circles of, 78 

centre of, 74 

conjugate diameters of, 86 

construction of, 76, 82, 83, 88, 
96, 97, 98 

diameters of, 74 

directrices of, 74 

focal properties of, 92, 93, 94 

foci of, 74, 76 

major and minor axes of, 74, 76 

mechanical method of describing 
an, 76 

mechanism for drawing an, 81 



Ellipse, principal axes of, 74, 87 
tangential properties of, 92, 93, 

94 

theorems on, 74, 75, 79, 80 
Elongation-load diagram, 166 
Embankments, contoured, 472 
Engine-divided scale, 2 
Envelope of a curve, 120 

of a tangent, 478 
Epicycloid, 114, 116 

tangent to, 116 

normal to, 116 
Epitrochoid, 114 
Equation to a curve, 154 

to a straight line, 160 
Equation, solution of cubic, 172 
Equator, 186 
Equiangular spiral, 126 
Equilibrant, 510, 514 
Equilibrium, 510 
Errors in drawing, 2, 51 

of observation, 170 
Euclid, 1, 50, 176 
Evolute of a curve, 122 
Evolution, 494 
Examination paper, 538 
Experiment, plotting results of, 166 
Extreme and mean ratio, 32 

Figured plans, 198, 275 

Focal properties of ellipse, 92, 93, 

94 

sphere, 72, 74, 322 
Focus of a conic, 72 
Force diagram, 518 

scale, 522 
Forces, concurrent, 510 

moment of, 520, 526 

parallel, 525, 526 

polygon of, 510 
Fourth proportional, 32, 488 
Frame, hinged, 516, 518, 521 
French curves, 3 
Friction of crane, 170 
Frustum of pyramid, 216 
Fundamental rules, 193-197 
Funicular polygon, 519, 524 

Generator, 70 



552 



PRACTICAL GEOMETRY 



Geometrical mean, 32 
Geometry, analytical, 156 

descriptive, 176 

practical, 1 

practical solid, 176 

pure, 1 

pure solid, 176 
Glass paper, 4 
Good draughtsmanship, 177 
Graphic arithmetic, 486 

statics, 502 
Graphical solution of equations, 172 
Gravity, centre of, 530 

Harmonic division of a line, 34 

mean, 34 

motion, 128, 132 

pencil, 34, 51 

progression, 34 

range, 34 
Helical spring, 476 

surface, 476 
Helix, pitch angle of, 476 

pitch of, 476 

projection of, 476 
Hinged frame, 516, 518, 521 
Horizontal projection, 274 
Hyperbola, 71, 104 

asymptotes of, 104 

centre of, 104 

properties of, 104 

rectangular, 154 

tangent and normal to, 106 

theorems on, 105 

transverse axis of, 104 

vertices of, 104 
Hypocycloid, 114, 116 

tangent and normal to, 116 
Hypotrochoid, 114 

Icosahedron, 463 
Ill-conditioned constructions, 10 
Important points, 385, 408 

problem, 219 

problem on shadows, 428 

sections, 396 
Inaccessible points, 30, 60 
Inclinations of line, 188, 206 

of plane, 191, 256 



Inclined plane, 214 

Inking-pen, 3 

Inscribed sphere, 320, 342 

Instantaneous centre, 88, 93, 112, 

122, 141 
Instruments, drawing, 1, 3 

accuracy of, 4 

setting of, 4 
Integral powers, 494 
Intercepts of line, 160 

of plane, 190 
Interpenetrations of solids, 384 
Interpolation, 169 
Intersection of cone and cylinder, 

390-395 
of cylinder and pyramid, 400 
of line and plane, 220, 224, 234, 

437 

of prism and pyramid, 398 

of shadows, 428 

of sphere and prism, 396 

of surfaces, 384 

of surfaces of revolution, 402 

of three planes, 250 

of two cylinders, 388 

of two planes, 250, 280 
Involute, centre of curvature of, 122 

of a curve, 122, 478 

tangent and normal to, 122 
Involution, 494 
Isometric axes and scale, 444 

projection, 444 

scale, construction of, 446 

Jib crane, 516 

Laboratory test of a crane,' 170 

Latitude, 187 

Law of crane, 173 

Length of circular arc, 112 

Limits of accuracy, 2 

Line and perpendicular plane, 204 

and plane, intersection of, 220, 
224, 234, 437 

division of, 10 

equation to straight, 160, 162 

inclinations of, 206 

of separation, 407, 416 

possible positions of, 204 



INDEX 



55$ 



Line, powers of, 494, 500 

projections of, 188 

square root of, 496 

to find equation to a, 162 

traces of, 188, 208 

unit, 486, 488 

width of a, 2 
Linear equation, 160 

laws, 173 

scale, 522 
Lines, product of, 488 

quotient of, 488 

shortest distance between, 288 

of slope, 224 
Link motion, 139, 174 

polygon, 519, 524 

polygon, properties of, 524 

polygon, closing of, 525, 532, 533 
Loaded string, 518 
Load-elongation diagram, 166 
Locus, 66, 84 
Logarithmic spiral, 124, 126 

curve, 126, 500 
Log of timber, volume of, 175 
Longitude, 186 

Major axis of an ellipse, 74 
Map, contour road-, 167 
Maps, contoured, 442 
Mathematical instruments, 3 
Maxima and minima, 169 
Mean and extreme ratio, 32 

geometrical, 32 

harmonic, 34 

ordinate, 44 

proportional, 32 
Measurements to scale, 177 
Mechanical appliances, 3, 4 

constraint, 136 

method of describing ellipse, 76 
Median, 56 
Meridian circle, 186 

plane, 186 
Method of sections, 384 
Metric axes, 443 

directions, 443 

lines, 443 

plane, 443 

projection, 442 



Metric scales, 443 
Mid-ordinate rule. 44 
Minor axis of ellipse, 74 
Miscellaneous problems, 462 
Model of auxiliary plane, 196 

of oblique plane, 190, 230 

of perpendicular plane, 214 

of planes of projection, 180, 193 

ol ruled surface, 479 
Models, use of, 177-191 
Moment diagram, 522, 526 

of a force, 520 

of a system of forces, 525 

partial, 525 

scale, 522 
Motions, component and resultant, 

^a. S4 
under mechanical constraint, 136 
Multiplication, graphic, 487 

Nodes, 118 

Non-concurrent forces, 520, 524 

resultant of, 524 
Normal to any curve, 90 

to a cone or cylinder, 548 

to a cycloid, 116 

to an ellipse, 92 

to an epicycloid, 116 

to a hyperbola, 106 

to a hypocycloid, 116 

to an involute, 122 

to a parabola, 101 

to a roulette, 112 
Notation, in solid geometry, 182 

Bow's, 514, 526 
Number, concrete, 486 

pure, 8, 486 

Oblique plane, 190, 214, 230 

converted into inclined plane, 
232 

model of, 190, 230 
Observation, correction of errors of, 
170 

plotting results of, 166 
Octahedron, 426 
Order or sequence, 503 
Ordinate, 44, 150 

mean, 44 



554 



PRACTICAL GEOMETRY 



Origin of co-ordinates, 150, 178 
Orthographic projection, 182 
of a conic, 78 

Pantograph, 26 
Paper,, drawing-, 3 
Paper, squared, 152 
Parabola, 71, 99 

construction of, 99, 102 

directrix of, 99 

focus of, 99 

normal to, 101 

properties of, 100 

tangent to, 101 

theorems on, 100 
Parallel curves, 120 

forces, 525, 526 

motion, 140 

projection, 182, 458 

rulers, 1 
Parallelogram of displacements, 507 

of forces, 511, 512 

of vectors, 506 
Partial moment, 525 

resultant, 524 
Pencil of rays, 34 

quality of, 3 

sharpening of, 4 
Period of vibration, 128 
Perpendicular plane, 214 

model of, 214 
Phase, 128 
Plan, 182 
Plane, 190 

inclinations of, 191, 256 

inclined, 214 

intercepts of, 190 

oblique, 190, 230 

perpendicular, 214 

rabatments of, 190, 219, 235, 
244 

scale of slope of, 224 

through three points, 248 

traces of, 190, 215, 231 
Planes, angle between, 249, 284 

front and side, 178 

horizontal and vertical, 178 

intersection of, 250, 280 

of projection, 182, 193, 195 



1 Planes of projection, models of, 180, 

193 

of reference, 178 
Plotting, examples on, 172 

a straight line, 162 

curves, 157 

points, 150 

results of experiment, 166 

results of observation, 166 
Point in space, 178, 180 

of inflexion, 118 

size of a, 2 

(x,y), 144 

(x, y, 2), 179 
Polar axis, 186 

co-ordinates, 149, 186 

triangle, 466 
Pole, 186, 519 

of spiral, 124 
Polygon, area of, 36, 492 

division of, 37 

funicular, 519 

link, 519 

of displacements, 508 

of forces, 510 

vector, 508 
Polygons, construction of regular, 52 

similar, 26 
Polyhedron, 462 

sphere circumscribed about, 464 

sphere inscribed in, 464 
Position in space, 176 

in space defined and exhibited, 
178, 180, 292 

of a point, 178, 180 

of a point in a plane, 148, 150 
Powers of a line, 494, 500 
Practical geometry, 1 

solid geometry, 176 
Pressure of steam, 175 
] Price chart, 167 
I Pricker, how to make a, 3 
j Problems, miscellaneous, 462 
Product of lines, 488 
Projection, definitions relating to, 
182 

isometric, 444 

horizontal, 274 

metric, 442, 458 



INDEX 



555 



Projection, oblique, 182, 458 

of cone, 320 

of curves, 308 

of cylinder, 320 

of a helix, 476 

of surface of revolution, 334 

of V-threaded screw, 476 

orthogonal or orthographic, 182 

parallel, 182 

planes of, 182, 193, 195 

sectional, 198 

trimetric, 454, 458 
Projections, figured, 198 
Projective properties of ellipse, 78 
Projectors, 182, 196 
Properties of conic sections, 72 

of cone and cylinder, 320 
Proportional, fourth, 32, 488 

mean, 32 

third, 32 
Proportion and ratio, 8 
Protractor, circular, 3 
Pure geometry, 1 

number, 8, 486 

solid geometry, 176 
Pyramid, development of, 216 

frustum of, 216 

Quadrilateral, complete, 34 
Quantity, directed, 502 

(r, 6, <p) co-ordinated, 186 
Rabatment, 190, 206, 218, 244 

model to show, 196, 245 

of a point, 196, 244 
Radial projection of a conic, 78 
Radius vector, 124 
Range, harmonic, 34 
Rankine's construction for length of 

circular arc, 112 
Ratio and proportion, 8 

representation by symbols, 8, 9 

theorems on, 8, 9 
Ratio, extreme and mean, 32 
Ratios, definition of trigonometrical, 

18, 19 
Rays, divergent, 407 

parallel, 407 

pencil of, 34 



Rectangular components of a vector, 
504. 512 
co-ordinates, 149, 150, 178 
hyperbola, 154 
Regular polyhedra, 462, 543 
Representation of vectors, 503 
Resistance-efficiency curve, 171 

-effort curve, 171 
Resolving vectors, 504, 512 
Resultant, 504, 512, 524 

motions, 130 

of non-concurrent forces, 524 

partial, 524 
Results of experiments, plotting of, 
166 

of observation, plotting of, 166 
Revolution, surface of, 334, 336, 

342, 422, 544 
Right-angled triangle, solution of, 

16 
Road-map, 167 
Rolling curves, no 
Roulettes, 1 10 

base of, 1 10 

centre of curvature of, 112 

construction of, no 

definition of, no 

normal to, 112 
Rule for area, 44 

mid-ordinate, 44 

ordinary, 44 

Simpson's, 44, 45 

Weddle's, 45 
Ruled celluloid, 10, 12 

surface, model of, 479 

surface, section of, 478 

tracing-paper, 10 
Rulers, parallel, 1 
Rules, fundamental, 193 

for adding vectors, 507, 508 

for mean ordinate, 44 

Scale, r, 6 
choice of, 168 
close and open divided, 6 
decimally divided, 6 
description of, 6 
diagonal, 14 
engine-divided, 2 



556 



PRACTICAL GEOMETRY 



Scale, examples on use of, 7 

force, 522 

isometric, 444, 446 

linear, 522 

metric, 443, 458 

moment, 522 

of chords, 22 

of slope of plane, 224 

of speed, 12 

trimetric, 443, 454 
Screw thread, 476 

pitch of, 476 

V-threaded, 476 
Sealing-wax, use of, 3 
Section I., plane geometry, 1 

II., solid geometry, 176 

III., graphics, 486 

of sphere by plane, 386 
Sectional projections, 198 
Sections, conic, 70 

method of, 384 
Separation, line ot, 407, 416** 
Sequence, or order, 503 
Set-squares, 1, 3 

adjustment of, 4 

errors in, 2 
Setting of instruments, 4 
Shadows of bolt, 424 

cast, 406 

of circle, line, point, 409-411 

geometrical, 406 

important problem on, 428 

intersection of, 428 

of rivet, 426 

of simple solids, 411-423 

theorems on, 407, 408, 424 
Sharpening of pencil, 4 
Shortest distance between two lines, 

212, 288 
Similar polygons, 26 
Simple harmonic motion, 128, 132 

amplitude of, 128 

period of, 128 

phase of, 128 
Sine curve, 128, 476 

definition of, 18 
Sines, table of, 20, 21 
Size of a point, 2 
Skeleton diagrams, 137 



Slide valve, 139, 169 
Slope, scale of, 224, 275 

line of, 224 
Solids in given positions, 292 
Solution of equations, 172 

of right-angled triangle, 16 
Space of three dimensions, 177 
Special curves, no 

points, 385 
Sphere inscribed in cone, 320 

in cylinder, 320 

section by plane, 386 
Spherical triangles, 466 
Spiral curves, 124 

equiangular, 126 

logarithmic, 124, 126 

of Archimedes, 124 

springs, 476 
Squared paper, 152 
Square root of line, 496 

of number, 496 
Statics, graphic, 502 
Steam pressure, 175 
Stephenson's link motion, 174 
Straight-edge, 2 

line, average position of, 170 

line, equation to, 

line, plotting of, 162 

line, projections of, 188, 205 
String, loaded, 518 
Subtraction, graphic, 487 
Summation, graphic, 487 
Surfaces, development of, 198, 
216, 470 

in contact, 368 

revolution, 334, 336, 342, 422 

ruled, 478 
System, concurrent, 510 
Systems of co-ordin?tes, 149, 180 

Tables of sines, cosines, etc., 20, 21 
Tangent, definition of, 19 

planes to surfaces, 340 

to any curve, 90 

to cycloid, 116 

to draw a, 50 

to ellipse, 92 

to epicycloid, 116 

to hyperbola, 106 



INDEX 



557 



Tangent to hypocycloid, 116 

to involute, 122 

to parabola, 101 
Tangential properties of cones, 255 

properties of ellipse, 92, 93, 94 
Tangents, table of, 20, 21 
Tee- square, 3 
Teeth, wheel, no, 120 
Temperature of steam, 175 
Templates, 3, in, 120, 138 

ruled, 10 
Testing machine, 166 
Test of crane, 170 
Tetrahedron, 464, 543 

height of, 302 
Theorems, 544 

on couples, 520 

on the ellipse, 74, 75, 79, 80, 
92 

on forces, 511, 524, 525 

on the hyperbola, 105 

on involutes and evolutes, 122 

on the line and circle, 51 

on metric projection, 459 

on the parabola, 100 

on shadows, 407, 408, 424 

on similar polygons, 26, 27, 42 
Third proportional, 32 
Thread used in plotting, 170 

screw, 476 
Three planes of reference, 178 

co-ordinate planes, 178 

metric axes, 443, 458 
Tie of crane, 516 
Tools, 2 

Tortuous curves, 384 
Trace of cone, 322, 326 

of cylinder, 324, 326 
Traces of a line, 188, 208 

of a plane, 190, 214, 231 
Tracing-paper, 170 

ruled, 10, 12 

template, 3, in, 120, 138 



Trammel for ellipse, 82, 83, 87 

triangular, 85, 87, 88 
Transversal, 34 
Triangle of displacements, 506 

of forces, 511, 513 

of vectors, 506 

polar, 466 

rabatment of, 190, 219 

spherical, 466 

solution of right-angled, 16 
Triangles, construction of, 52-59 
Triangular trammel, 85, 87, 88 
Trigonometrical tables, 20, 21 
Trihedral angles, 179, 466 
Trimetric axes, 443, 454, 458 

scales, 454, 458 
True alignment, 2 
Trying plane, 2, 4 

! Unit line, 486, 488 
Useful construction, 243 
Use o*f models, 177, 191 
of squared paper, 152 

Vector, definition of, 502 

equality, 508, 512 

parallelogram, 506 

polygon, 508 

radius, 124 

representation of, 503 

summation, 508 

triangle, 506 
Vectorally equal, 508, 512 
Vectorial angle, 124 
Vertical plane, 214 
Volume of log of timber, 175 

of 1 lb. of steam, 175 
V-threaded screw, 476 

Watt's parallel motion, 140 
Ways of defining angles, 18 
Wheel teeth, no, 120 
Width of a line, 2 



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